[LeetCode]House Robber II

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After robbing those houses on that street, the thief has found himself a new place for his thievery so that he will not get too much attention. This time, all houses at this place are arranged in a circle. That means the first house is the neighbor of the last one. Meanwhile, the security system for these houses remain the same as for those in the previous street.

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.

[思路]这次是对环处理。我们可以把这个问题变换成在直线上处理的子问题。见子问题解法。

因为是环，所以最后一个和第一个不能同时取。

那么这个问题，我们对第一个到倒数第二个求在直线上的解，再对第二个到最后一个求在直线上的解，取最大值即可。

class Solution {
public:
    int rob(vector<int>& nums) {
        if(nums.size()==1)
            return nums[0];
        return max(RobLine(nums,0,nums.size()-1),RobLine(nums,1,nums.size()));
    }
    int RobLine(vector<int> &num,int start,int end ) {  
        if(num.size()==0)  
            return 0;  
        int step1=0;  
        int step2=0;  
        int step3=num[start];  
        int step4=0;  
        for(int i=start+1;i<end;++i){  
            step4=max(num[i]+step2,num[i-1]+step1);  
            step1=step2;  
            step2=step3;  
            step3=step4;  
        }  
        return step3;  
    }  
};

[LeetCode]House Robber II

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原文地址：http://blog.csdn.net/ciaoliang/article/details/46523623