day04-算法热题10题

LeetCode 热题 Hot 100

• 75. 颜色分类

class Solution {
    public void sortColors(int[] nums) {
        int lastZero = -1, firstTwo = nums.length;
        int curr = 0;
        while (curr < firstTwo) {
            // 红色
            if (nums[curr] == 0) {
                // 与lastZero的下一位进行交换
                swap(nums, curr, lastZero + 1);
                lastZero++;
                   // 防止curr是0
                curr++;
            }
            // 白色
            else if (nums[curr] == 1) {
                curr++;
            } else {
                // 与firstTwo的前一位进行交换
                swap(nums, curr, firstTwo - 1);
                firstTwo--;
            }
        }
    }


    private void swap(int[] nums, int i, int j) {
        if (i == j) return;
        nums[i] = nums[i] ^ nums[j];
        nums[j] = nums[i] ^ nums[j];
        nums[i] = nums[i] ^ nums[j];
    }

}

• 76. 最小覆盖子串

class Solution {
    /**
     * 滑动窗口
     */
    public String minWindow(String s, String t) {
        // 异常判断
        if (s == null || t == null || s.length() == 0 || t.length() == 0) return "";
        // 用来存储t中每个字符出现的次数
        Map<Character, Integer> ht = new HashMap<>();
        // 用来当前窗口中每个字符出现的次数
        Map<Character, Integer> hs = new HashMap<>();
        // 遍历进行记录
        for (char c : t.toCharArray()) {
            ht.put(c, ht.getOrDefault(c, 0) + 1);
        }
        String answer = "";
        int minLen = Integer.MAX_VALUE;
        int n = s.length();
        // 如果出现过了就移动
        int end = 0;
        // 记录匹配了几次
        int matched = 0;
        for (int start = 0; start < n; start++) {
            // 不是每一个字符都匹配上
            while (end < n && matched < ht.size()) {
                char ch = s.charAt(end);
                hs.put(ch, hs.getOrDefault(ch, 0) + 1);
                // 如果出现的次数是一样的代表匹配上了
                if (hs.get(ch).equals(ht.getOrDefault(ch, 0))) {
                    matched++;
                }
                // 没有匹配上,右指针往右边移动一
                end++;
            }
            // 如果匹配完t中所有出现的字符之后
            if (matched == ht.size()) {
                if (end - start < minLen) {
                    minLen = end - start;
                    answer = s.substring(start, end);
                }
            }
            // 左指针移动回去
            char ch = s.charAt(start);
            hs.put(ch, hs.get(ch) - 1);
            if (hs.get(ch).equals(ht.getOrDefault(ch, 0) - 1)) {
                matched--;
            }
        }
        return answer;
    }
}

• 78. 子集

class Solution {
  LinkedList<List<Integer>> pathList = new LinkedList<>();
    LinkedList<Integer> path = new LinkedList<>();

    public List<List<Integer>> subsets(int[] nums) {
        backtrack(nums, 0);
        return pathList;
    }

    private void backtrack(int[] nums, int start) {
        pathList.add(new LinkedList<>(path));
        while (start < nums.length) {
            path.add(nums[start]);
            backtrack(nums, start + 1);
            path.removeLast();
            start++;
        }
    }
}

• 79. 单词搜索

    int m;
    int n;
    int w;
    char[] letters;
    char[][] board;
    boolean[][] visited;

    /**
     * 回溯法
     */
    public boolean exist(char[][] board, String word) {
        this.m = board.length;
        this.n = board[0].length;
        this.w = word.length();
        this.letters = word.toCharArray();
        this.board = board;
        this.visited = new boolean[m][n];
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                // 0 是word里的index
                boolean res = backtrack(i, j, 0);
                if (res) return true;
            }
        }
        return false;
    }

    private boolean backtrack(int i, int j, int wordIndex) {
        if (wordIndex >= w) return true;
        if (i < 0 || j < 0 || i >= m || j >= n || letters[wordIndex] != board[i][j] || visited[i][j]) return false;
        // 访问过了就不能再去看了
        visited[i][j] = true;
        boolean res = backtrack(i + 1, j, wordIndex + 1) || backtrack(i, j + 1, wordIndex + 1)
                ||
                backtrack(i - 1, j, wordIndex + 1) || backtrack(i, j - 1, wordIndex + 1);
        visited[i][j] = false;
        return res;
    }

• 84. 柱状图中最大的矩形

• 85. 最大矩形

• 94. 二叉树的中序遍历

 /**
 * 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 {
  /**
     * 递归
     * 判断左子树和右子树是否是对称的
     * 左边右子树=右边的左子树&&左边的左子树=右边的右子树
     */
    public boolean isSymmetric(TreeNode root) {
        if (root == null) return true;
        // 检查两棵子树是否对称
        return check(root.left, root.right);
    }

    private boolean check(TreeNode left, TreeNode right) {
        if (left == null && right == null) return true;
        if (left == null || right == null) return false;
        // 两个根节点需要相同
        if (left.val != right.val) return false;
        // 左右子节点需要对称相同
        return check(left.right, right.left) && check(left.left, right.right);
    }
}

• 96. 不同的二叉搜索树

class Solution {
  /**
     * 1 ~ n
     * [1,i-1] i [i+1,n]
     * f(n) = sum( f(i-1) * f(n-i) )
     */
    public int numTrees(int n) {
        int[] dp = new int[n + 1];
        dp[0] = 1;
        for (int i = 1; i <= n; i++) {
            for (int j = 0; j <= i - 1; j++) {
                dp[i] += dp[j] * dp[i - 1 - j ];
            }
        }
        return dp[n];
    }
}

• 98. 验证二叉搜索树

/**
 * 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 {
   /**
     * 节点的左子树只包含 小于 当前节点的数。
     * 节点的右子树只包含 大于 当前节点的数。
     * 所有左子树和右子树自身必须也是二叉搜索树。
     */
    public boolean isValidBST(TreeNode root) {
        return isValid(root, Long.MIN_VALUE, Long.MAX_VALUE);
    }

    private boolean isValid(TreeNode root, long minValue, long maxValue) {
        if (root == null) return true;
        if (root.left != null) {
            // 左节点的值必须大于最小值,小于根节点
            if (root.left.val > minValue && root.left.val < root.val) {
                boolean valid = isValid(root.left, minValue, root.val);
                if (!valid) return false;
            } else {
                return false;
            }
        }
        if (root.right != null) {
            if (root.right.val < maxValue && root.right.val > root.val) {
                boolean valid = isValid(root.right, root.val, maxValue);
                if (!valid) return false;
            } else {
                return false;
            }
        }
        return true;
    }
}

/**
 * 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 boolean isValidBST(TreeNode root) {
        return isValid(root, null, null);
    }
    // 以 root 为根的子树节点必须满足 max.val > root.val > min.val
     private boolean isValid(TreeNode root, TreeNode min, TreeNode max) {
        if (root == null) return true;
        if (min != null && min.val >= root.val) return false;
        if (max != null && max.val <= root.val) return false;
        return isValid(root.left, min, root) && isValid(root.right, root, max);
    }
}

• 101. 对称二叉树

/**
 * 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 {
  /**
     * 递归
     * 判断左子树和右子树是否是对称的
     * 左边右子树=右边的左子树&&左边的左子树=右边的右子树
     */
    public boolean isSymmetric(TreeNode root) {
        if (root == null) return true;
        // 检查两棵子树是否对称
        return check(root.left, root.right);
    }

    private boolean check(TreeNode left, TreeNode right) {
        if (left == null && right == null) return true;
        if (left == null || right == null) return false;
        // 两个根节点需要相同
        if (left.val != right.val) return false;
        // 左右子节点需要对称相同
        return check(left.right, right.left) && check(left.left, right.right);
    }
}