213. House Robber II

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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.

Credits:
Special thanks to @Freezen for adding this problem and creating all test cases.

public class Solution {
    public int rob(int[] nums) {
        //偷最后一家和不偷最后一家
        if(nums.length == 0) return 0;
        if(nums.length == 1) return nums[0];
        int n = nums.length;
        int[] dp1 = new int[n];
        int[] dp2 = new int[n];
     
        dp1[0] = nums[0];
        dp1[1] = Math.max(nums[1], nums[0]);
        for(int i = 2 ; i < n-1 ; i++){
            dp1[i] = Math.max(dp1[i-1] , dp1[i-2] + nums[i]);
        }
        
        dp2[n-1] = nums[n-1];
        dp2[n-2] = Math.max(nums[n-1], nums[n-2]);
        for(int i = n-3 ; i > 0 ; i--){
            dp2[i] = Math.max(dp2[i+1] , dp2[i+2] + nums[i]);
        }
        return Math.max(dp2[1] , dp1[n-2]);
    }
}

213. House Robber II

标签：list   pre   eating   discuss   edits   elf   ext   and   without   

原文地址：http://www.cnblogs.com/joannacode/p/6108089.html