LeetCode每日一题JAVA、JavaSrcipt题解——2022.08.21-08.31

---

title: 2022-08-21-1455-检查单词是否为句中其他单词的前缀
date: 2022-08-21 15:01:12
tags: [Daily Practice, 简单题, 字符串, 双指针, 模拟]
categories: 算法题

---

1455. 检查单词是否为句中其他单词的前缀

模拟

JAVA

class Solution {
    public int isPrefixOfWord(String sentence, String searchWord) {
        // 切割成单词
        String [] ch = sentence.split(" ");
        for (int i = 0; i < ch.length; i++) {
            if (isqz(searchWord, ch[i])) {
                return i + 1;
            }
        }
        return -1;
    }
    // 判断A是否为a的前缀
    public boolean isqz(String a, String A) {
        int la = a.length(), lA = A.length();
        if (la > lA) {
            return false;
        }
        for (int i = 0; i < la; i++) {
            if (a.charAt(i) != A.charAt(i)) {
                return false;
            }
        }
        return true;
    }
}

JS

/**
 * @param {string} sentence
 * @param {string} searchWord
 * @return {number}
 */
var isPrefixOfWord = function(sentence, searchWord) {
    // 切割成单词
    let ch = sentence.split(" ");
    for (let i = 0; i < ch.length; i++) {
        if (isqz(searchWord, ch[i])) {
            return i + 1;
        }
    }
    return -1;
};
// 判断A是否为a的前缀
const isqz = function (a, A) {
    let la = a.length, lA = A.length;
    if (la > lA) {
        return false;
    }
    for (let i = 0; i < la; i++) {
        if (a.charAt(i) !== A.charAt(i)) {
            return false;
        }
    }
    return true;
}

双指针

JAVA

class Solution {
    public int isPrefixOfWord(String sentence, String searchWord) {
        int n = sentence.length(), index = 1, start = 0, end = 0;
        while (start < n) {
            while (end < n && sentence.charAt(end) != ' ') {
                end++;
            }
            if (isPrefix(sentence, start, end, searchWord)) {
                return index;
            }
            index++;
            end++;
            start = end;
        }
        return -1;
    }

    public boolean isPrefix(String sentence, int start, int end, String searchWord) {
        for (int i = 0; i < searchWord.length(); i++) {
            if (start + i >= end || sentence.charAt(start + i) != searchWord.charAt(i)) {
                return false;
            }
        }
        return true;
    }
}

JS

var isPrefixOfWord = function(sentence, searchWord) {
    let n = sentence.length, index = 1, start = 0, end = 0;
    while (start < n) {
        while (end < n && sentence[end] !== ' ') {
            end++;
        }
        if (isPrefix(sentence, start, end, searchWord)) {
            return index;
        }

        index++;
        end++;
        start = end;
    }
    return -1;
}

const isPrefix = (sentence, start, end, searchWord) => {
    for (let i = 0; i < searchWord.length; i++) {
        if (start + i >= end || sentence[start + i] !== searchWord[i]) {
            return false;
        }
    }
    return true;
};

---

title: 2022-08-22-655-输出二叉树
date: 2022-08-22 19:01:12
tags: [Daily Practice, 中等题, dfs]
categories: 算法题

---

655. 输出二叉树

dfs

JAVA

/**
 * 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 {
    private List<List<String>> ans;
    private int height;
    public List<List<String>> printTree(TreeNode root) {
        height = dfs1(root);
        ans = new ArrayList<>(height);
        int n = (1 << height) - 1;
        for (int i = 0; i < height; i++) {
            List<String> list = new ArrayList<>(n);
            for (int j = 0; j < n; j++) {
                list.add("");
            }
            ans.add(list);
        }
        dfs2(root, 0, (n - 1) >> 1);
        return ans;
    }

    private int dfs1(TreeNode node) {
        return node != null ? Math.max(dfs1(node.left), dfs1(node.right)) + 1 : 0;
    }

    private void dfs2(TreeNode node, int r, int c) {
        ans.get(r).set(c, node.val + "");
        int diff = 1 << height - 2 - r;
        if (node.left != null) {
            dfs2(node.left, r + 1, c - diff);
        }
        if (node.right != null) {
            dfs2(node.right, r + 1, c + diff);
        }
    }
}

JS

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {string[][]}
 */
var printTree = function(root) {
    const calDepth = (root) => {
        let h = 0;
        if (root.left) {
            h = Math.max(h, calDepth(root.left) + 1);
        }
        if (root.right) {
            h = Math.max(h, calDepth(root.right) + 1);
        }
        return h;
    }

    const dfs = (res, root, r, c, height) => {
        res[r][c] = root.val.toString();
        if (root.left) {
            dfs(res, root.left, r + 1, c - (1 << (height - r - 1)), height);
        }
        if (root.right) {
            dfs(res, root.right, r + 1, c + (1 << (height - r - 1)), height);
        }
    }

    const height = calDepth(root);
    const m = height + 1;
    const n = (1 << (height + 1)) - 1;
    const res = new Array(m).fill(0).map(() => new Array(n).fill(''));
    dfs(res, root, 0, Math.floor((n - 1) / 2), height);
    return res;
};

---

title: 2022-08-23-782-变为棋盘
date: 2022-08-23 22:01:12
tags: [Daily Practice, 困难题]
categories: 算法题

---

782. 变为棋盘

JAVA

class Solution {
    public int movesToChessboard(int[][] board) {
        int n = board.length;
        int rowMask = 0, colMask = 0;        

        /* 检查棋盘的第一行与第一列 */
        for (int i = 0; i < n; i++) {
            rowMask |= (board[0][i] << i);
            colMask |= (board[i][0] << i);
        }
        int reverseRowMask = ((1 << n) - 1) ^ rowMask;
        int reverseColMask = ((1 << n) - 1) ^ colMask;
        int rowCnt = 0, colCnt = 0;
        for (int i = 0; i < n; i++) {
            int currRowMask = 0;
            int currColMask = 0;
            for (int j = 0; j < n; j++) {
                currRowMask |= (board[i][j] << j);
                currColMask |= (board[j][i] << j);
            }
            /* 检测每一行的状态是否合法 */
            if (currRowMask != rowMask && currRowMask != reverseRowMask) {
                return -1;
            } else if (currRowMask == rowMask) {
                /* 记录与第一行相同的行数 */
                rowCnt++;
            }
            /* 检测每一列的状态是否合法 */
            if (currColMask != colMask && currColMask != reverseColMask) {
                return -1;
            } else if (currColMask == colMask) {
                /* 记录与第一列相同的列数 */
                colCnt++;
            }
        }
        int rowMoves = getMoves(rowMask, rowCnt, n);
        int colMoves = getMoves(colMask, colCnt, n);
        return (rowMoves == -1 || colMoves == -1) ? -1 : (rowMoves + colMoves); 
    }

    public int getMoves(int mask, int count, int n) {
        int ones = Integer.bitCount(mask);
        if ((n & 1) == 1) {
            /* 如果 n 为奇数，则每一行中 1 与 0 的数目相差为 1，且满足相邻行交替 */
            if (Math.abs(n - 2 * ones) != 1 || Math.abs(n - 2 * count) != 1 ) {
                return -1;
            }
            if (ones == (n >> 1)) {
                /* 以 0 为开头的最小交换次数 */
                return n / 2 - Integer.bitCount(mask & 0xAAAAAAAA);
            } else {
                return (n + 1) / 2 - Integer.bitCount(mask & 0x55555555);
            }
        } else { 
            /* 如果 n 为偶数，则每一行中 1 与 0 的数目相等，且满足相邻行交替 */
            if (ones != (n >> 1) || count != (n >> 1)) {
                return -1;
            }
            /* 找到行的最小交换次数 */
            int count0 = n / 2 - Integer.bitCount(mask & 0xAAAAAAAA);
            int count1 = n / 2 - Integer.bitCount(mask & 0x55555555);  
            return Math.min(count0, count1);
        }
    }
}

---

title: 2022-08-24-1460-通过翻转子数组使两个数组相等
date: 2022-08-24 19:01:12
tags: [Daily Practice, 简单题]
categories: 算法题

---

1460. 通过翻转子数组使两个数组相等

排序

JAVA

class Solution {
    public boolean canBeEqual(int[] target, int[] arr) {
        Arrays.sort(target);
        Arrays.sort(arr);
        return Arrays.equals(target, arr);
    }
}

JS

var canBeEqual = function(target, arr) {
    target.sort((a, b) => a - b);
    arr.sort((a, b) => a - b);
    return target.toString() === arr.toString();
};

哈希

JAVA

class Solution {
    public boolean canBeEqual(int[] target, int[] arr) {
        Map<Integer, Integer> counts1 = new HashMap<Integer, Integer>();
        Map<Integer, Integer> counts2 = new HashMap<Integer, Integer>();
        for (int num : target) {
            counts1.put(num, counts1.getOrDefault(num, 0) + 1);
        }
        for (int num : arr) {
            counts2.put(num, counts2.getOrDefault(num, 0) + 1);
        }
        if (counts1.size() != counts2.size()) {
            return false;
        }
        for (Map.Entry<Integer, Integer> entry : counts1.entrySet()) {
            int key = entry.getKey(), value = entry.getValue();
            if (!counts2.containsKey(key) || counts2.get(key) != value) {
                return false;
            }
        }
        return true;
    }
}

JS

var canBeEqual = function(target, arr) {
    const counts1 = new Map();
    const counts2 = new Map();
    for (const num of target) {
        counts1.set(num, (counts1.get(num) || 0) + 1);
    }
    for (const num of arr) {
        counts2.set(num, (counts2.get(num) || 0) + 1);   
    }
    if (counts1.size !== counts2.size) {
        return false;
    }
    for (const [key, value] of counts1.entries()) {
        if (!counts2.has(key) || counts2.get(key) !== value) {
            return false;
        }
    }
    return true;
};

---

title: 2022-08-25-658-找到 K 个最接近的元素
date: 2022-08-25 23:01:12
tags: [Daily Practice, 中等题, 二分, 排序]
categories: 算法题

---

658. 找到 K 个最接近的元素

排序

直接按照要求做差并排序再比较选取

JAVA

class Solution {
    public List<Integer> findClosestElements(int[] arr, int k, int x) {
        List<Integer> list = new ArrayList<Integer>();
        for (int num : arr) {
            list.add(num);
        }
        Collections.sort(list, (a, b) -> {
            if (Math.abs(a - x) != Math.abs(b - x)) {
                return Math.abs(a - x) - Math.abs(b - x);
            } else {
                return a - b;
            }
        });
        List<Integer> ans = list.subList(0, k);
        Collections.sort(ans);
        return ans;
    }
}

JS

var findClosestElements = function(arr, k, x) {
    const list = [...arr];
    list.sort((a, b) => {
        if (Math.abs(a - x) !== Math.abs(b - x)) {
            return Math.abs(a - x) - Math.abs(b - x);
        } else {
            return a - b;
        }
    });
    const ans = list.slice(0, k);
    ans.sort((a, b) => a - b);
    return ans;
};

二分

JAVA

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

    public int binarySearch(int[] arr, int x) {
        int low = 0, high = arr.length - 1;
        while (low < high) {
            int mid = low + (high - low) / 2;
            if (arr[mid] >= x) {
                high = mid;
            } else {
                low = mid + 1;
            }
        }
        return low;
    }
}

JS

var findClosestElements = function(arr, k, x) {
    let right = binarySearch(arr, x);
    let left = right - 1;
    while (k-- > 0) {
        if (left < 0) {
            right++;
        } else if (right >= arr.length) {
            left--;
        } else if (x - arr[left] <= arr[right] - x) {
            left--;
        } else {
            right++;
        }
    }
    const ans = [];
    for (let i = left + 1; i < right; i++) {
        ans.push(arr[i]);
    }
    return ans;
}

const binarySearch = (arr, x) => {
    let low = 0, high = arr.length - 1;
    while (low < high) {
        const mid = low + Math.floor((high - low) / 2);
        if (arr[mid] >= x) {
            high = mid;
        } else {
            low = mid + 1;
        }
    }
    return low;
}

---

title: 2022-08-26-1464-数组中两元素的最大乘积
date: 2022-08-26 18:01:12
tags: [Daily Practice, 简单题, 排序]
categories: 算法题

---

1464. 数组中两元素的最大乘积

排序

JAVA

class Solution {
    public int maxProduct(int[] nums) {
        int len = nums.length;
        Arrays.sort(nums);
        return (nums[len - 1] - 1) * (nums[len - 2] - 1);
    }
}

JS

/**
 * @param {number[]} nums
 * @return {number}
 */
var maxProduct = function(nums) {
    let len = nums.length;
    nums.sort((a, b) => a - b);
    return (nums[len - 1] - 1) * (nums[len - 2] - 1);
};

// 官解
var maxProduct = function(nums) {
    nums.sort((a, b) => a - b);
    console.log(nums)
    return (nums[nums.length - 1] - 1) * (nums[nums.length - 2] - 1);
};

一次遍历，维护最大和次大值

JAVA

class Solution {
    public int maxProduct(int[] nums) {
        int a = nums[0], b = nums[1];
        if (a < b) {
            int temp = a;
            a = b;
            b = temp;
        }
        for (int i = 2; i < nums.length; i++) {
            if (nums[i] > a) {
                b = a;
                a = nums[i];
            } else if (nums[i] > b) {
                b = nums[i];
            }
        }
        return (a - 1) * (b - 1);
    }
}

JS

var maxProduct = function(nums) {
    let a = nums[0], b = nums[1];
    if (a < b) {
        const temp = a;
        a = b;
        b = temp;
    }
    for (let i = 2; i < nums.length; i++) {
        if (nums[i] > a) {
            b = a;
            a = nums[i];
        } else if (nums[i] > b) {
            b = nums[i];
        }
    }
    return (a - 1) * (b - 1);
};

---

title: 2022-08-27-662-二叉树最大宽度
date: 2022-08-27 13:01:12
tags: [Daily Practice, 中等题, 二叉树, dfs, bfs]
categories: 算法题

---

662. 二叉树最大宽度

bfs

JAVA

class Solution {
    public int widthOfBinaryTree(TreeNode root) {
        int res = 1;
        List<Pair<TreeNode, Integer>> arr = new ArrayList<Pair<TreeNode, Integer>>();
        arr.add(new Pair<TreeNode, Integer>(root, 1));
        while (!arr.isEmpty()) {
            List<Pair<TreeNode, Integer>> tmp = new ArrayList<Pair<TreeNode, Integer>>();
            for (Pair<TreeNode, Integer> pair : arr) {
                TreeNode node = pair.getKey();
                int index = pair.getValue();
                if (node.left != null) {
                    tmp.add(new Pair<TreeNode, Integer>(node.left, index * 2));
                }
                if (node.right != null) {
                    tmp.add(new Pair<TreeNode, Integer>(node.right, index * 2 + 1));
                }
            }
            res = Math.max(res, arr.get(arr.size() - 1).getValue() - arr.get(0).getValue() + 1);
            arr = tmp;
        }
        return res;
    }
}

JS

dfs

JAVA

class Solution {
    Map<Integer, Integer> levelMin = new HashMap<Integer, Integer>();

    public int widthOfBinaryTree(TreeNode root) {
        return dfs(root, 1, 1);
    }

    public int dfs(TreeNode node, int depth, int index) {
        if (node == null) {
            return 0;
        }
        levelMin.putIfAbsent(depth, index); // 每一层最先访问到的节点会是最左边的节点，即每一层编号的最小值
        return Math.max(index - levelMin.get(depth) + 1, Math.max(dfs(node.left, depth + 1, index * 2), dfs(node.right, depth + 1, index * 2 + 1)));
    }
}

JS

var widthOfBinaryTree = function(root) {
    /** JS 存在计数溢出的问题，使用 BigInt，BigInt 不能调用 Math 中的方法。 */
    let maxWidth = 1n;
    const leftIds = []
    const dfs = (root, level, currIdx) => {
        if (leftIds[level] === undefined) {
            leftIds[level] = currIdx;
        } else {
            const width = currIdx - leftIds[level] + 1n;
            maxWidth = maxWidth > width ? maxWidth : width;
        }
        if (root.left !== null) {
            dfs(root.left, level + 1, currIdx * 2n - 1n);
        }
        if (root.right !== null) {
            dfs(root.right, level + 1, currIdx * 2n);
        }
    }
    dfs(root, 0, 1n);
    return maxWidth;    
};

---

title: 2022-08-28-793-阶乘函数后K个零
date: 2022-08-28 13:01:12
tags: [Daily Practice, 困难题, 数学, 二分]
categories: 算法题

---

793. 阶乘函数后 K 个零

数学 + 二分

JAVA

// 数学
class Solution {
    public int preimageSizeFZF(int k) {
        //确定阶梯值范围 最终的到的K < start
        int start = 1;
        while (start < K){
            start = start * 5+1;
        }
        //确定范围后，执行精确查找
        while (start > 1){
            //只有5以下阶乘才会出现start-1成立，其它情况不会存在，因为任何一个阶段分界值都会包含一个以上的5
            if(start - 1 == K){
                //不存在的返回0
                return 0;
            }
            //逆推下一个阶梯值 从f(x+1) 推导出f(x)
            start = (start - 1) / 5;
            //获取剩余值，进行下一阶梯运算
            K %= start;
        }
        //只要存在，必然是5个
        return 5;
    }
}

// 递归 + 二分
class Solution {
    public int preimageSizeFZF(int k) {
        long l = k - 1,r = k * 10L + 1;
        while(l + 1 < r){
            long m = l + r >> 1,t = f(m);
            if(t == k) return 5;
            else if(t < k) l = m;
            else r = m;
        }
        return 0;
    }
    long f(long n){
        if(n == 0) return 0;
        return n / 5 + f(n / 5);
    }
}

// 三叶 数学 + 二分
class Solution {
    public int preimageSizeFZF(int k) {
        if (k <= 1) return 5;
        return f(k) - f(k - 1);
    }
    int f(int x) {
        long l = 0, r = (long) 1e10;
        while (l < r) {
            long mid = l + r + 1 >> 1;
            if (getCnt(mid) <= x) l = mid;
            else r = mid - 1;
        }
        return (int)r;
    }
    long getCnt(long x) {
        long ans = 0;
        while (x != 0) {
            ans += x / 5; x /= 5;
        }
        return ans;
    }
}

JS

var preimageSizeFZF = function(k) {
    return help(k + 1) - help(k);
}

const help = (k) => {
    let r = 5 * k;
    let l = 0;
    while (l <= r) {
        const mid = Math.floor((l + r) / 2);
        if (zeta(mid) < k) {
            l = mid + 1;
        } else {
            r = mid - 1;
        }
    }
    return r + 1;
}

const zeta = (x) => {
    let res = 0;
    while (x != 0) {
        res += Math.floor(x / 5);
        x = Math.floor(x / 5);
    }
    return res;
};

---

title: 2022-08-29-1470-重新排列数组
date: 2022-08-29 9:01:12
tags: [Daily Practice, 简单题, 模拟]
categories: 算法题

---

1470. 重新排列数组

一次遍历

JAVA

class Solution {
    public int[] shuffle(int[] nums, int n) {
        int [] cnt = new int[2 * n];
        for (int i = 0, j = 0; i < 2 * n; i++, j++) {
            cnt[i++] = nums[j];
            cnt[i] = nums[j + n];
        }
        return cnt;
    }
}

// 官解
class Solution {
    public int[] shuffle(int[] nums, int n) {
        int[] ans = new int[2 * n];
        for (int i = 0; i < n; i++) {
            ans[2 * i] = nums[i];
            ans[2 * i + 1] = nums[i + n];
        }
        return ans;
    }
}

JS

var shuffle = function(nums, n) {
    var cnt = new Array(2 * n).fill(0);
    for (let i = 0, j = 0; i < 2 * n; i++, j++) {
        cnt[i++] = nums[j];
        cnt[i] = nums[j + n];
    }
    return cnt;
};

// 官解
var shuffle = function(nums, n) {
    const ans = new Array(2 * n).fill(0);
    for (let i = 0; i < n; i++) {
        ans[2 * i] = nums[i];
        ans[2 * i + 1] = nums[i + n];
    }
    return ans;
};

---

title: 2022-08-30-998-最大二叉树 II
date: 2022-08-30 22:01:12
tags: [Daily Practice, 中等题, 二叉树]
categories: 算法题

---

998. 最大二叉树 II

JAVA

/**
 * 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 TreeNode insertIntoMaxTree(TreeNode root, int val) {
        TreeNode parent = null;
        TreeNode cur = root;
        while (cur != null) {
            if (val > cur.val) {
                if (parent == null) {
                    return new TreeNode(val, root, null);
                }
                TreeNode node = new TreeNode(val, cur, null);
                parent.right = node;
                return root;
            } else {
                parent = cur;
                cur = cur.right;
            }
        }
        parent.right = new TreeNode(val);
        return root;
    }
}

JS

var insertIntoMaxTree = function(root, val) {
    let parent = null;
    let cur = root;
    while (cur) {
        if (val > cur.val) {
            if (!parent) {
                return new TreeNode(val, root, null);
            }
            let node = new TreeNode(val, cur, null);
            parent.right = node;
            return root;
        } else {
            parent = cur;
            cur = cur.right;
        }
    }
    parent.right = new TreeNode(val);
    return root;
};

---

title: 2022-08-30-946-验证栈序列
date: 2022-08-31 15:01:12
tags: [Daily Practice, 中等题, 栈, 模拟]
categories: 算法题

---

946. 验证栈序列

模拟

JAVA

class Solution {
    public boolean validateStackSequences(int[] pushed, int[] popped) {
        Deque<Integer> stack = new ArrayDeque<>();
        int j = 0; //索引popped
        for (int i = 0; i < pushed.length; i++) {
            stack.push(pushed[i]);
            while (!stack.isEmpty() && stack.peek() == popped[j]) {
                j++;
                stack.pop();
            }
        }
        return stack.isEmpty();
    }
}

JS

/**
 * @param {number[]} pushed
 * @param {number[]} popped
 * @return {boolean}
 */
var validateStackSequences = function(pushed, popped) {
    let stack = [];
        let j = 0;
        for (let i = 0, l = pushed.length; i < l; i++) {
            stack.push(pushed[i]);
            while (stack.length != 0 && stack[stack.length - 1] == popped[j]) {
                j++;
                stack.pop();
            }
        }
    return stack.length === 0;
};

双指针

JAVA

class Solution {
    public boolean validateStackSequences(int[] pushed, int[] popped) {
        int n = pushed.length, idx = 0;
        for (int i = 0, j = 0; i < n; i++) {
            pushed[idx++] = pushed[i];
            while (idx > 0 && pushed[idx - 1] == popped[j] && ++j >= 0) idx--;
        }
        return idx == 0;
    }
}

JS

/**
 * @param {number[]} pushed
 * @param {number[]} popped
 * @return {boolean}
 */
var validateStackSequences = function(pushed, popped) {
    let n = pushed.length, idx = 0;
    for (let i = 0, j = 0; i < n; i++) {
        pushed[idx++] = pushed[i];
        while (idx > 0 && pushed[idx - 1] == popped[j] && ++j >= 0) idx--;
    }
    return idx == 0;
};