广度优先 (Breadth first search,

发布时间:2024年01月15日

广度优先 (Breadth first search, BFS)

// 计算从起点 start 到终点 target 的最近距离
int BFS(Node start, Node target) {
    Queue<Node> q; // 核心数据结构
    Set<Node> visited; // 避免走回头路
    
    q.offer(start); // 将起点加入队列
    visited.add(start);

    while (q not empty) {
        int sz = q.size();
        /* 将当前队列中的所有节点向四周扩散 */
        for (int i = 0; i < sz; i++) {
            Node cur = q.poll();
            /* 划重点:这里判断是否到达终点 */
            if (cur is target)
                return step;
            /* 将 cur 的相邻节点加入队列 */
            for (Node x : cur.adj()) {
                if (x not in visited) {
                    q.offer(x);
                    visited.add(x);
                }
            }
        }
    }
    // 如果走到这里,说明在图中没有找到目标节点
}

二叉树的最小深度

给定一个二叉树,找出其最小深度。

最小深度是从根节点到最近叶子节点的最短路径上的节点数量。

//leetcode submit region begin(Prohibit modification and deletion)

import java.util.LinkedList;
import java.util.Queue;

//
//*
// * Definition for a binary tree node.
// * public class TreeNode {
// *     int val;
// *     TreeNode left;
// *     TreeNode right;
// *     TreeNode() {}
// *     TreeNode(int val) { this.val = val; }
// *     TreeNode(int val, TreeNode left, TreeNode right) {
// *         this.val = val;
// *         this.left = left;
// *         this.right = right;
// *     }
// * }
// */
class Solution {

    public int minDepth(TreeNode root) {
        //bfs广度优先,得出的肯定是最小深度
        if (root == null) {
            return 0;
        }
        Queue<TreeNode> q = new LinkedList<>();
        q.offer(root);

        int step = 1;
        while (!q.isEmpty()) {

            int sz = q.size();

            for (int i = 0; i < sz; i++) {
                TreeNode cur = q.poll();
                if (cur.left == null && cur.right == null) {
                    return step;
                }

                if (cur.left != null) {
                    q.offer(cur.left);
                }
                if (cur.right != null) {
                    q.offer(cur.right);
                }
            }
            step++;
        }
        return -1;
    }
}
//leetcode submit region end(Prohibit modification and deletion)

打开转盘锁,采用双向bfs

import java.util.*;


//leetcode submit region begin(Prohibit modification and deletion)
class Solution {
    public int openLock(String[] deadends, String target) {
        Set<String> q1 = new HashSet<>();
        Set<String> q2 = new HashSet<>();
        Set<String> dead = new HashSet<>(Arrays.asList(deadends));
        Set<String> visited = new HashSet<>();

        q1.add("0000");
        q2.add(target);

        int step = 0;
        while (!q1.isEmpty() && !q2.isEmpty()) {

            Set<String> temp = new HashSet<>();
            for (String current : q1) {
                if (dead.contains(current)) {
                    continue;
                }
                if (q2.contains(current)) {
                    return step;
                }
                visited.add(current);

                for (int i = 0; i < 4; i++) {
                    String upString = up(current, i);
                    if (!visited.contains(upString)) {
                        temp.add(upString);
                    }

                    String downString = down(current, i);
                    if (!visited.contains(downString)) {
                        temp.add(downString);
                    }
                }
            }
            q1 = q2;
            q2 = temp;
            step++;
        }

        return -1;
    }

    public String up(String source, int j) {
        char ch[] = source.toCharArray();
        if (ch[j] == '9') {
            ch[j] = '0';
        }
        else {
            ch[j] += 1;
        }
        return new String(ch);
    }

    public String down(String source, int j) {
        char ch[] = source.toCharArray();
        if (ch[j] == '0') {
            ch[j] = '9';
        }
        else {
            ch[j] -= 1;
        }
        return new String(ch);
    }

}
//leetcode submit region end(Prohibit modification and deletion)

文章来源:https://blog.csdn.net/qq_44290178/article/details/135606479
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