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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.
class Solution { public: int rob(vector<int> &num) { int n = num.size(); if(n == 0) return 0; else if(n == 1) return num[0]; else { vector<int> maxV(n, 0); maxV[0] = num[0]; maxV[1] = max(num[0], num[1]); for(int i = 2; i < n; i ++) maxV[i] = max(maxV[i-2]+num[i], maxV[i-1]); return maxV[n-1]; } } };
int max(int a, int b) { if(a > b) return a; else return b; } int rob(int* nums, int numsSize) { int best0 = 0; // 表示没有选择当前houses int best1 = 0; // 表示选择了当前houses for(int i = 0; i < numsSize; i++){ int temp = best0; best0 = max(best0, best1); // 没有选择当前houses，那么它等于上次选择了或没选择的最大值 best1 = temp + nums[i]; // 选择了当前houses，值只能等于上次没选择的+当前houses的money } return max(best0, best1); }
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原文地址：http://www.cnblogs.com/dylqt/p/4945992.html
