Zoe Ding's Blog

恢复写博客的第一周：写点什么？

2026-09-13T00:00:00Z

上周把博客恢复之后，我决定把写博客这件事恢复起来。

重新写博客

2026-09-07T00:00:00Z

上一篇博文是8年前了。时间过的真快。过去这8年，发生了太多太多。

今天用Claude Code 把这个博客简单整理了一下。然后，我决定重新开始写博客了。

1197. Minimum Knight Moves

2022-11-20T00:00:00Z

1197. Minimum Knight Moves

BFS. Note: Using the HashSet to keep track of the visited cells in the Java implementation will result in the TLE (Time Limit Exceeded) exception.

class Solution {
    public int minKnightMoves(int x, int y) {
        Queue<int[]> queue = new LinkedList<>();
        queue.offer(new int[]{0, 0});
        int[][] dirs = `{`{1, 2}, {2, 1}, {-2, 1}, {-1, 2}, {2, -1}, {1, -2}, {-1, -2}, {-2, -1}`}`;
        boolean[][] visited = new boolean[602][602];
        
        int moves = 0;
        while(!queue.isEmpty()) {
            int size = queue.size();
            for (int i = 0; i < size; i++) {
                int[] cur = queue.poll();
                if (cur[0] == x && cur[1] == y) {
                    return moves;
                }
                for (int[] dir : dirs) {
                    int row = cur[0] + dir[0];
                    int col = cur[1] + dir[1];
                    if (row < -300 || row > 300 || col < -300 || col > 300 || visited[row + 301][col + 301]) {
                        continue;
                    }
                    queue.offer(new int[]{row, col});
                    visited[row + 301][col + 301] = true;
                }
            }
            moves ++;
        }
        return -1;
        
    }
}

658. Find K Closest Elements

2022-11-20T00:00:00Z

658. Find K Closest Elements

Binary Search. 提前一步停下来。

class Solution {
    public List<Integer> findClosestElements(int[] arr, int k, int x) {
        List<Integer> res = new ArrayList<>();
        int closest = findClosest(arr, x);
        
        int left = closest;
        int right = closest;
        while (k > 1) {
            if (left == 0) {
                right ++;
            } else if (right == arr.length - 1) {
                left --;
            } else if (arr[right + 1] - x < x - arr[left - 1]) {
                right ++;
            } else {
                left --;
            }
            k--;
        }
        for (int i = left; i <= right; i++) {
            res.add(arr[i]);
        }
        
        return res;
    }
    
    private int findClosest(int[] arr, int x) {
        int left = 0;
        int right = arr.length - 1;
        while (left < right - 1) {
            int mid = left + (right - left) / 2;
            if (arr[mid] == x) {
                return mid;
            } else if (arr[mid] > x) {
                right = mid;
            } else {
                left = mid;
            }
        }
        if (arr[right] - x >= x - arr[left]) {
            return left;
        } else {
            return right;
        }            
    }
}

353. Design Snake Game

2022-08-23T00:00:00Z

2296. Design a Text Editor

用 Deque 做。加 Set 可以优化时间，但是加 Set 跑不通不知道哪里有问题。。。

class SnakeGame {
    
    // Set<Integer> set;
    Deque<Integer> deque;
    int width;
    int height;
    int[][] foods;
    int foodIndex;

    public SnakeGame(int width, int height, int[][] food) {
        this.width = width;
        this.height = height;
        this.foods = food;
        this.foodIndex = 0;
        // this.set = new HashSet<>();
        // set.add(0); 
        deque = new ArrayDeque<>();
        deque.offerLast(0);
    }
    
    public int move(String direction) {

        int row = deque.peekFirst() / width;
        int col = deque.peekFirst() % width;
        if (direction.equals("U")) {
            row --;
        } else if (direction.equals("D")) {
            row ++;
        } else if (direction.equals("L")) {
            col --;
        } else { // direction.equals("R")
            col ++;
        }
        int newHead = row * width + col;
        int tail = deque.peekLast();
        if (newHead != tail && deque.contains(newHead)) {
            return -1;
        }
        if (row < 0 || row >= height || col < 0 || col >= width) {
            return -1;
        }
        deque.offerFirst(newHead);
        // set.add(newHead);
        if (foodIndex < foods.length && row == foods[foodIndex][0] && col == foods[foodIndex][1]) {
            foodIndex ++;
        } else {
            int cur = deque.pollLast();
            // set.remove(cur);
        }
        return foodIndex;
    }
}

/**
 * Your SnakeGame object will be instantiated and called as such:
 * SnakeGame obj = new SnakeGame(width, height, food);
 * int param_1 = obj.move(direction);
 */

2296. Design a Text Editor

2022-08-22T00:00:00Z

2296. Design a Text Editor

Use StringBuilder.

class TextEditor {
    StringBuilder sb;
    int index;

    public TextEditor() {
        // 0 1 2 3 4 5 6
        // h e l l | o 
        sb = new StringBuilder();
        index = 0;
    }
    
    public void addText(String text) {
        sb.insert(index, text);
        index += text.length();
    }
    
    public int deleteText(int k) {
        int tmp = index;
        index -= k;
        if (index < 0) {
            index = 0;
        }
        sb.delete(index, tmp);
        return tmp - index;
    }
    
    public String cursorLeft(int k) {
        index -= k;
        if (index < 0) {
            index = 0;
        }
        if (index < 10) {
            return sb.substring(0, index);
        } else {
            return sb.substring(index - 10, index);
        }
    }
    
    public String cursorRight(int k) {
        index += k;
        if (index > sb.length()) {
            index = sb.length();
        }
        if (index < 10) {
            return sb.substring(0, index);
        } else {
            return sb.substring(index - 10, index);
        }
    }
}

/**
 * Your TextEditor object will be instantiated and called as such:
 * TextEditor obj = new TextEditor();
 * obj.addText(text);
 * int param_2 = obj.deleteText(k);
 * String param_3 = obj.cursorLeft(k);
 * String param_4 = obj.cursorRight(k);
 */

787. Cheapest Flights Within K Stops

2022-08-19T00:00:00Z

787. Cheapest Flights Within K Stops

Sol1. BFS (TLE)

class Solution {
    public int findCheapestPrice(int n, int[][] flights, int src, int dst, int k) {
        Map<Integer, List<int[]>> map = new HashMap<>();
        for (int[] flight : flights) {
            map.put(flight[0], map.getOrDefault(flight[0], new ArrayList<>()));
            map.get(flight[0]).add(new int[]{flight[1], flight[2]});
        }
        int step = 0;
        Queue<int[]> queue = new LinkedList<>();
        int res = Integer.MAX_VALUE;
        queue.offer(new int[]{src, 0});
        while (!queue.isEmpty()) {
            int size = queue.size();
            for (int i = 0; i < size; i++) {
                int[] cur = queue.poll();
                if (cur[0] == dst) {
                    res = Math.min(res, cur[1]);
                }
                if (!map.containsKey(cur[0])) {
                    continue;
                }
                for (int[] flight : map.get(cur[0])) {
                    if (cur[1] + flight[1] > res) {
                        continue; // 强剪枝
                    }
                    queue.offer(new int[]{flight[0], cur[1] + flight[1]});
                }
            }
            if (step > k) {
                break;
            }
            step ++;
        }
        return res == Integer.MAX_VALUE ? -1 :res;
    }
}

Sol2. Dijkstra's Algorithm

Dijkstra's algorithm. Dijkstra 是贪心算法。

参考了这个帖子

class Solution {
    public int findCheapestPrice(int n, int[][] flights, int src, int dst, int k) {
        Map<Integer, List<int[]>> map = new HashMap<>();
        for (int[] flight : flights) {
            map.put(flight[0], map.getOrDefault(flight[0], new ArrayList<>()));
            map.get(flight[0]).add(new int[]{flight[1], flight[2]});
        }
        // Maintain a PQ to fetch the min cost with max K stops. PQ will have 3 items - 
        //  1. node - The node to visit
        //  2. cost - Cost to visit the node
        //  3. stops - Remaining number of stops to visit destinatio
        PriorityQueue<int[]> queue = new PriorityQueue<>((o1, o2) -> o1[1] - o2[1]);
        queue.offer(new int[]{src, 0, 0});
        
        // Maintains the minimum number of stops from a particular city. 
        Map<Integer, Integer> seen = new HashMap<>();
        while (!queue.isEmpty()) {
            int[] c = queue.poll();
            int city = c[0];
            int cost = c[1];
            int stops = c[2];
            if (!seen.containsKey(city) || stops < seen.get(city)) {
                seen.put(city, stops);
            } else {
                continue;
            }
            if (city == dst) {
                return cost;
            }
            if (stops > k || !map.containsKey(city)) {
                continue;
            }

            for (int[] next : map.get(city)) {
                queue.add(new int[]{next[0], cost + next[1], stops + 1});
            }
        }
        return -1;
    }
}

Sol3. Bellman-Ford Algorithm

Bellman-Ford Algorithm. Bellman-Ford Algorithm 是动态规划算法。

class Solution {
    public int findCheapestPrice(int n, int[][] flights, int src, int dst, int k) {
        int[] cost = new int[n];
        Arrays.fill(cost,Integer.MAX_VALUE);
        cost[src] = 0;
        for(int i = 0; i <= k; i++)
        {
            int[] temp = Arrays.copyOf(cost,n);
            for(int[] f: flights) {
                int curr = f[0];
                int next = f[1];
                int price = f[2];
                if (cost[curr] == Integer.MAX_VALUE) {
                    continue;
                }
                temp[next] = Math.min(temp[next], cost[curr] + price);
            }
            cost = temp;
        }
        return cost[dst] == Integer.MAX_VALUE ? -1 : cost[dst];
    }
}

1298. Maximum Candies You Can Get from Boxes

2022-08-18T00:00:00Z

1298. Maximum Candies You Can Get from Boxes

class Solution {
    public int maxCandies(int[] status, int[] candies, int[][] keys, int[][] containedBoxes, int[] initialBoxes) {
        Set<Integer> keySet = new HashSet<>();
        Set<Integer> boxesToOpenNoKey = new HashSet<>();
        Queue<Integer> queue = new LinkedList<>();
        for (int box : initialBoxes) {
            queue.offer(box);
        }
        Set<Integer> visitedBoxes = new HashSet<>();
        int totalCandies = 0;
        while (!queue.isEmpty()) {
            int size = queue.size();
            for (int i = 0; i < size; i++) {
                int curBox = queue.poll();
                if (visitedBoxes.contains(curBox)) {
                    continue;
                }
                int curStatus = status[curBox];
                int curCandies = candies[curBox];
                int[] curKeys = keys[curBox];
                for (int key : curKeys) {
                    keySet.add(key);
                    if (boxesToOpenNoKey.contains(key)) {
                        totalCandies += candies[key];
                        boxesToOpenNoKey.remove(key);
                    }
                }
                
                if (curStatus == 1) {
                    totalCandies += curCandies;
                } else {
                    if (keySet.contains(curBox)) {
                        totalCandies += curCandies;
                    } else {
                        boxesToOpenNoKey.add(curBox);
                    }
                }
                
                visitedBoxes.add(curBox);
                int[] curContainedBoxes = containedBoxes[curBox];
                for (int box : curContainedBoxes) {
                    queue.offer(box);
                }
            }
        }
        return totalCandies;
    }
}

815. Bus Routes

2022-08-17T00:00:00Z

767. Reorganize String

BFS. Use two sets to store visited routes and stops.

class Solution {
    public int numBusesToDestination(int[][] routes, int source, int target) {
        Map<Integer, List<Integer>> stopToRouteMap = new HashMap<>();
        for (int i = 0; i < routes.length; i++) {
            for (int j = 0; j < routes[i].length; j++) {                
                if (!stopToRouteMap.containsKey(routes[i][j])) {
                    stopToRouteMap.put(routes[i][j], new ArrayList<>());
                }
                stopToRouteMap.get(routes[i][j]).add(i);
            }
        }
        if (!stopToRouteMap.containsKey(source)) {
            return -1;
        }
        // [[1,2,7],[3,6,7]]
        // 1: 0
        // 2: 0
        // 3: 1
        // 6: 2
        // 7: 1,2
        Queue<Integer> queue = new LinkedList<>();
        queue.offer(source);
        Set<Integer> stopVisited = new HashSet<>();
        Set<Integer> routeVisited = new HashSet<>();
        stopVisited.add(source);
        int numOfBuses = 0;
        while (!queue.isEmpty()) {
            int size = queue.size();
            for (int i = 0; i < size; i++) {
                int curStop = queue.poll();
                if (curStop == target) {
                    return numOfBuses;
                }
                if (stopToRouteMap.containsKey(curStop)) {
                    List<Integer> curRoutes = stopToRouteMap.get(curStop);
                    for (int route : curRoutes) {
                        if (routeVisited.contains(route)) {
                            continue;
                        }
                        routeVisited.add(route);
                        int[] stops = routes[route];
                        for (int stop : stops) {
                            if (!stopVisited.contains(stop)) {
                                queue.offer(stop);
                                stopVisited.add(stop);
                            }
                        }
                    }
                }
            }
            numOfBuses ++;
        }
        return -1;
    }
}

767. Reorganize String

2022-08-15T00:00:00Z

767. Reorganize String

Hashmap + PriorityQueue.

class Solution {
    public String reorganizeString(String s) {
        Map<Character, Integer> map = new HashMap<>();
        for (char c : s.toCharArray()) {
            map.put(c, map.getOrDefault(c, 0) + 1);
        }
        PriorityQueue<Character> pq = new PriorityQueue((n1, n2) -> map.get(n2) - map.get(n1));
        pq.addAll(map.keySet());
        StringBuilder sb = new StringBuilder();
        while (pq.size() > 1) {
            char c1 = pq.poll();
            char c2 = pq.poll();
            sb.append(c1);
            sb.append(c2);
            map.put(c1, map.get(c1) - 1);
            map.put(c2, map.get(c2) - 1);
            if (map.get(c1) > 0) {
                pq.offer(c1);
            }
            if (map.get(c2) > 0) {
                pq.offer(c2);
            }
        }
        while (!pq.isEmpty()) {
            char c = pq.remove();
            if (map.get(c) > 1) {
                return "";
            }
            sb.append(c);
        }
        return sb.toString();
    }
}