leetcode 213 : House Robber II

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House Robber II

Note: This is an extension of House Robber.

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]

House Robber I的升级版. 因为第一个element 和最后一个element不能同时出现. 则分两次call House Robber I. case 1: 不包括最后一个element. case 2: 不包括第一个element.

两者的最大值即为全局最大值

[CODE]

public class Solution {
    //1 2 3
    public int rob(int[] nums) {
        if(nums==null || nums.length==0) return 0;
        if(nums.length==1) return nums[0];
        if(nums.length==2) return Math.max(nums[0], nums[1]);
        return Math.max(robsub(nums, 0, nums.length-2), robsub(nums, 1, nums.length-1));
    }
    
    private int robsub(int[] nums, int s, int e) {
        int n = e - s + 1;
        int[] d =new int[n];
        d[0] = nums[s];
        d[1] = Math.max(nums[s], nums[s+1]);
        
        for(int i=2; i<n; i++) {
            d[i] = Math.max(d[i-2]+nums[s+i], d[i-1]);
        }
        return d[n-1];
    }
}

leetcode 213 : House Robber II

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原文地址：http://blog.csdn.net/xudli/article/details/45886721