今日任务

198.打家劫舍
213.打家劫舍II
337.打家劫舍 III

198.打家劫舍

题目链接： . - 力扣（LeetCode）

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

213.打家劫舍II

题目链接： . - 力扣（LeetCode）

class Solution {
    public int rob(int[] nums) {
        if (nums.length == 0) return 0;
        if (nums.length == 1) return nums[0];
        return Math.max(robotRange(nums, 0, nums.length - 2), robotRange(nums, 1, nums.length - 1));
    }
    int robotRange(int[] nums, int start, int end) {
        if (start == end) return nums[start];
        int[] dp = new int[nums.length];
        dp[start] = nums[start];
        dp[start + 1] = Math.max(nums[start], nums[start + 1]);
        for (int i = start + 2; i <= end ; i++) {
            dp[i] = Math.max(dp[i - 2] + nums[i], dp[i - 1]);
        }
        return dp[end];
    }
}

337.打家劫舍 III

题目链接： . - 力扣（LeetCode）

/**
 * 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 rob(TreeNode root) {
        int[] result = robTree(root);
        return Math.max(result[0], result[1]);
    }
    // 0-不偷 1-偷
    int[] robTree(TreeNode root) {
        if (root == null) return new int[]{0, 0};
        int[] left = robTree(root.left);
        int[] right = robTree(root.right);

        //偷当前节点，那就不能偷左右节点
        int val1 = root.val + left[0] + right[0];

        //不偷当前节点，那左右节点可投可不投，取大值
        int val2 = Math.max(left[0], left[1]) + Math.max(right[0], right[1]);
        return new int[]{val2, val1};
    }    
}