力扣热题 100

文章目录

- 哈希
- 双指针
- 滑动窗口
- 子串
- 普通数组
- 矩阵
- 链表
- 二叉树
- 图论
- 回溯
- 二分查找
- 栈
- 堆
- 贪心算法
- 动态规划
- 多维动态规划
- 技巧

哈希

双指针

移动零

class Solution {
    public void moveZeroes(int[] nums) {
        int k = 0;
        for(int i = 0;i < nums.length; i++){
            if(nums[i] != 0) {
                nums[k] = nums[i];
                k++;
            }
        }
        for (int i = k; i < nums.length; i++) {
            nums[i] = 0;
        }
    }
}

class Solution {
    public void moveZeroes(int[] nums) {
        int k = 0;
        for (int i = 0; i < nums.length; i++) {
            if (nums[i] != 0) {
                int temp = nums[i];
                nums[i] = nums[k];
                nums[k] = temp;
                k++;
            }
        }
    }
}

盛最多水的容器

class Solution {
    public int maxArea(int[] height) {
        int left = 0, right = height.length - 1;
        int ans = 0;
        while (left < right) {
            int area = Math.min(height[left], height[right]) * (right - left);
            ans = Math.max(ans, area);
            if (height[left] <= height[right]) {
                left++;
            } else {
                right--;
            }
        }
        return ans;
    }
}

接雨水

class Solution {
    public int trap(int[] height) {
        int left = 0, right = height.length - 1;
        int leftMax = 0, rightMax = 0;

        int ans = 0;
        while (left < right) {
            leftMax = Math.max(leftMax, height[left]);
            rightMax = Math.max(rightMax, height[right]);
            if (height[left] < height[right]) {
                ans += leftMax - height[left];
                left++;
            } else {
                ans += rightMax - height[right];
                right--;
            }
        }
        return ans;
    }
}

滑动窗口

子串

普通数组

矩阵

链表

二叉树

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 {
    public List<Integer> inorderTraversal(TreeNode root) {
        List<Integer> reusltList = new ArrayList<>();
        inorderVisited(root, reusltList);
        return  reusltList;
    }

    private void inorderVisited(TreeNode root, List<Integer> reusltList) {
        if (root == null) {
            return;
        }
        inorderVisited(root.left, reusltList);
        reusltList.add(root.val);
        inorderVisited(root.right, reusltList);
    }
}

class Solution {
    public List<Integer> inorderTraversal(TreeNode root) {
        List<Integer> reusltList = new ArrayList<>();
        inorderVisited(root, reusltList);
        return  reusltList;
    }

    private void inorderVisited(TreeNode root, List<Integer> reusltList) {
        Stack<TreeNode> stack = new Stack<>();
        while (root != null || !stack.empty()) {
            while (root != null) {
                stack.push(root);
                root = root.left;
            }
            root = stack.pop();
            reusltList.add(root.val);
            root = root.right;
        }
    }
}

图论

回溯

二分查找

栈

堆

贪心算法

动态规划

在这里插入图片描述

爬楼梯

class Solution {
    public int climbStairs(int n) {
        int[] dp = new int[n+1];
        dp[0] = 1;
        dp[1] = 1;
        for(int i = 2;i <= n; i++){
            dp[i] = dp[i-1] + dp[i-2];
        }
        return dp[n];
    }
}

打家劫舍

class Solution {
    public int rob(int[] nums) {
        if (nums.length == 0) {
            return 0;
        }

        int len = nums.length;
        int[] dp = new int[len+1];
        dp[0] = 0;
        dp[1] = nums[0];
        for(int i = 2; i <= len; i++){
            dp[i] = Math.max(dp[i-1], dp[i-2] + nums[i-1]);
        }
        return dp[len];
    }
}

完全平方数

class Solution {
    public int numSquares(int n) {
        int[] dp = new int[n + 1];
        for(int i = 1; i <= n;i++) {
            int minn = Integer.MAX_VALUE;
            for(int j = 1; j * j <= i; j++){
                minn = Math.min(minn, dp[i - j * j]);
            }
            dp[i] = minn + 1;
        }
        return dp[n];
    }
}

零钱兑换

class Solution {
    public int coinChange(int[] coins, int amount) {
        int[] dp = new int[amount + 1];
        Arrays.fill(dp, amount + 1);
        dp[0] = 0;
        for (int i = 1; i <= amount; i++) {
            for (int coin : coins) {
                if (i < coin) {
                    continue;
                }
                dp[i] = Math.min(dp[i], dp[i - coin]);
            }
            dp[i] += 1;
        }
        return dp[amount] > amount ? -1 : dp[amount];
    }
}

class Solution {
    public int coinChange(int[] coins, int amount) {
        int[] dp = new int[amount + 1];
        Arrays.fill(dp, amount + 1);
        dp[0] = 0;
        for (int i = 1; i <= amount; i++) {
            for (int coin : coins) {
                if (i < coin) {
                    continue;
                }
                dp[i] = Math.min(dp[i], dp[i - coin] + 1);
            }
        }
        return (dp[amount] == amount + 1) ? -1 : dp[amount];
    }
}

分割等和子集

class Solution {
    public boolean canPartition(int[] nums) {
        if (nums == null || nums.length == 0) {
            return false;
        }
        int len = nums.length;
        int sum = 0;
        for (int num : nums) {
            sum += num;
        }
        if (sum % 2 != 0)
            return false;
        int target = sum / 2;
        int[] dp = new int[target + 1];
        for (int i = 0; i < len; i++) {
            for (int j = target; j >= nums[i]; j--) {
                dp[j] = Math.max(dp[j], dp[j - nums[i]] + nums[i]);
            }

            if (dp[target] == target) {
                return true;
            }
        }
        return dp[target] == target;
    }
}

多维动态规划

技巧

只出现一次的数字

class Solution {
    public int singleNumber(int[] nums) {
        int ans = 0;
        for (int num : nums) {
            ans ^= num;
        }
        return ans;
    }
}