198. 打家劫舍

題目

鏈接

代碼

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

213. 打家劫舍 II

題目

鏈接

代碼

class Solution {public int rob(int[] nums) {if(nums.length==1){return nums[0];}if(nums.length==2){return Math.max(nums[0],nums[1]);}int[] dpLeft = new int[nums.length];//不偷最后一個(gè)int[] dpRight = new int[nums.length];//不偷第一個(gè)dpLeft[0] = nums[0];dpLeft[1] = Math.max(nums[0],nums[1]);dpRight[1] = nums[1];if(nums.length>=3){dpRight[2] = Math.max(nums[1],nums[2]);}for(int i =2;i<nums.length-1;i++){//不偷最后一個(gè)dpLeft[i] = Math.max(dpLeft[i-1],dpLeft[i-2]+nums[i]);}for(int i =3;i<nums.length;i++){//不偷前一個(gè)dpRight[i] = Math.max(dpRight[i-1],dpRight[i-2]+nums[i]);}return Math.max(dpLeft[nums.length-2],dpRight[nums.length-1]);}
}

337. 打家劫舍 III

題目

鏈接

代碼

/*** 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 {Map<TreeNode,Integer> map = new HashMap();public int rob(TreeNode root) {if(root==null){return 0;}if(map.containsKey(root)){return map.get(root);}int money = root.val;if(root.left!=null){money+=rob(root.left.left)+rob(root.left.right);}if(root.right!=null){money+=rob(root.right.left)+rob(root.right.right);}int res = Math.max(money,rob(root.left)+rob(root.right));map.put(root,res);return res;}
}

121. 買賣股票的最佳時(shí)機(jī)

題目

鏈接

代碼

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