Leetcode 题目解析之 House Robber
原创Leetcode 题目解析之 House Robber
原创
ruochen
发布于 2022-02-14 13:19:01
发布于 2022-02-14 13:19:01
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.
本质：求数组不相邻元素最大和
动态规划（背包问题）：设Pi表示从0~i个房间抢劫的最大收益。
每次迭代只需要P的两个元素，并不需要设数组P。设两个变量为：
take :nums[i] + P[i-2]
nonTake:P[i-1] public int rob(int[] nums) {
int take = 0;
int nonTake = 0;
int max = 0;
for (int i = 0; i < nums.length; i++) {
take = nums[i] + nonTake;
nonTake = max;
max = Math.max(take, nonTake);
}
return max;
}
public int rob2(int[] nums) {
if (nums.length == 0) {
return 0;
}
if (nums.length == 1) {
return nums[0];
}
int[] P = new int[nums.length];
P[0] = nums[0];
P[1] = Math.max(nums[0], nums[1]);
for (int i = 2; i < nums.length; i++) {
P[i] = Math.max(nums[i] + P[i - 2], P[i - 1]);
}
return P[nums.length - 1];
}原创声明：本文系作者授权腾讯云开发者社区发表，未经许可，不得转载。
如有侵权，请联系 cloudcommunity@tencent.com 删除。
原创声明：本文系作者授权腾讯云开发者社区发表，未经许可，不得转载。
如有侵权，请联系 cloudcommunity@tencent.com 删除。
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