LeetCode-198. House Robber (Java)

You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed, the only constraint stopping you from robbing each of them is that adjacent houses have security system connected and it will automatically contact the police if two adjacent houses were broken into on the same night.

Given a list of non-negative integers representing the amount of money of each house, determine the maximum amount of money you can rob tonight without alerting the police.

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题意

求数组中元素相加后的最大值，要求在计算时相邻的两个元素只能加一个。

思路

首先如果数组长度为空的话，就返回0；

如果数组长度为1的话，返回nums[0];

如果数组长度为2的话，返回Math.max(nums[0],nums[1])；

如果数组长度大于2的话：使用动态规划，所以要是整体得到最大值，那么局部就要得到最大值，这个局部

就指从三个元素中得到最大值。因为题目限定，所以不可能是两个相邻的元素相加，只能是两个相邻的元素选择一个，

最大值要么是a+c，或者就是b。

状态：res[i]，表示第i天偷盗的最大值。

状态转移方程： res[i] = Math.max(res[i-2]+nums[i], res[i]);

代码

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