代码随想录35期Day47-Java

Day47题目

LeetCode198.打家劫舍1

核心思想:dp[i]=Math.max(dp[i-1],dp[i-2]+nums[i]),每一项是前一项或者前前一项+自身的最大值

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

LeetCode213打家劫舍2

核心思想:这里需要从第一个到倒数第二个之间和第二个到最后一个,这两种情况用打家劫舍1的方法计算结果,然后取其中最大的

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 res1 = robRoom(nums,0,nums.length-2);
        int res2 = robRoom(nums,1,nums.length-1);
        return Math.max(res1,res2);
    }
    public int robRoom(int[] nums,int start, int end){
        if (end == start) return nums[start];
        int[] dp = new int[nums.length+1];
        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-1],dp[i-2]+nums[i]);
        }
        return dp[end];
    } 
}

LeetCode337打家劫舍3

核心思想:变成了树,使用的dp数组很巧妙,只需要计算选他的最大值和不选的最大值.这点很难想到

/**
 * 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[] res = robRoom(root);
        return res[0]> res[1] ? res[0] : res[1];
    }
    public int[] robRoom(TreeNode node){
        int[] res = new int[2];
        if(node == null){
            return res;
        }
        int[] left = robRoom(node.left);
        int[] right = robRoom(node.right);
        // 计算选当前节点的最大值和不选的最大值
        res[0] = Math.max(left[0],left[1]) + Math.max(right[1],right[0] );
        res[1] = left[0]+right[0]+node.val;
        return res;
    }
}