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leetcode198 - House Robber - easy

时间:2020-10-27 10:55:44      阅读:21      评论:0      收藏:0      [点我收藏+]

标签:问题   index   each   ann   call   mount   security   bin   question   

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.

 

Example 1:

Input: nums = [1,2,3,1]
Output: 4
Explanation: Rob house 1 (money = 1) and then rob house 3 (money = 3).
             Total amount you can rob = 1 + 3 = 4.

Example 2:

Input: nums = [2,7,9,3,1]
Output: 12
Explanation: Rob house 1 (money = 2), rob house 3 (money = 9) and rob house 5 (money = 1).
             Total amount you can rob = 2 + 9 + 1 = 12.

 

Constraints:

  • 0 <= nums.length <= 100
  • 0 <= nums[i] <= 400
 
求取数列中不相互相邻的数的最大值,极值问题考虑dp- dp[i]表示抢到i为止最多抢了多少。
初始化:dp[0] = nums[0], dp[1] = max(nums[0], nums[1])
f:对于当前这户,要么不抢,那总共抢的钱就和dp[i-1]一样;抢的话,说明i-1没得抢,那总和是dp[i-1]+nums[i],每次取这两者较大值。
这里index i指向的是第i+1户人家,最后返还dp[n-1]
 
实现:Time O(n), Space O(1)
class Solution {
public:
    int rob(vector<int>& nums) {
        
        if (nums.empty()) return 0;
        int n = nums.size();
        if (n == 1) return nums[0];
        if (n == 2) return max(nums[0], nums[1]);
        
        vector<int> dp(n);
        dp[0] = nums[0];
        dp[1] = max(nums[0], nums[1]);
        for (int i=2; i<n; i++){
            dp[i] = max(dp[i-1], dp[i-2]+nums[i]);
        }
        return dp[n-1];
        
    }
};

 

 

leetcode198 - House Robber - easy

标签:问题   index   each   ann   call   mount   security   bin   question   

原文地址:https://www.cnblogs.com/xuningwang/p/13877908.html

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