题目详情
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.题目的意思是,我们是一个江洋大盗~现在我们要去偷整条街的房子,每个房子里有一定的钱。但是任何临近的两个房子被偷就会触发警报。要求我们求出在不触发警报的情况下偷到的最多的钱。每个房子里的钱通过输入的int数组表示。
想法
- 动态规划问题
- 对于每一个房子,我们有偷/不偷两种选择
- 因此我们声明两个变量prevNo和prevYes分别保存,我没偷/偷了当前房子的情况下,目前为止偷的最多的钱数。
- 如果想偷当前房子,那么要求我们并没有偷前一个房子,所以用前一个房子的prevNo值和当前房子的钱数相加。
- 如果不偷当前房子,那我们可以取前一个房子的prevNo值和prevYes值中较大的那个。
解法
public int rob(int[] nums) {
int prevNo = 0;
int prevYes = 0;
for(int n: nums){
int temp = prevNo;
prevNo = Math.max(prevNo, prevYes);
prevYes = temp+n;
}
return Math.max(prevNo, prevYes);
}
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