Day 48 动态规划part09

今日任务

• 198.打家劫舍
• 213.打家劫舍II
• 337.打家劫舍III

代码实现

基础打家劫舍

class Solution {public static int rob(int[] nums) {if (nums == null || nums.length == 0) return 0;if (nums.length == 1) return nums[0];int[] dp = new int[nums.length];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];}
}

环形打家劫舍

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

二叉树打家劫舍，难想

/*** 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[] ints = robAction(root);return Math.max(ints[0], ints[1]);}public int[] robAction(TreeNode root) {//dp[0]表示不加本节点的最大值 dp[1]表示加上本节点的最大值int[] dp = new int[2];if (root == null) return dp;int[] left = robAction(root.left);int[] right = robAction(root.right);//不加本节点，下一个节点加不加都行dp[0] = Math.max(left[0], left[1]) + Math.max(right[0], right[1]);dp[1] = root.val + left[0] + right[0];return dp;}}

今日总结

1. 基础思路了解后，每一个特殊思路还是比较难想，理解为主吧
2. 今天又大跌，5000+下跌，跌跌不休啊