leetcode 213 : House Robber II，leetcoderobber

leetcode 213 : House Robber II，leetcoderobber
House Robber IITotal Accepted: 654 Total Submissions: 2631

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 arearranged 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 tonightwithout alerting the police.

[思路]

House Robber I的升級版. 因為第一個element 和最後一個element不能同時出現. 則分兩次call House Robber I. case 1: 不包括最後一個element. case 2: 不包括第一個element.

兩者的最大值即為全域最大值

[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];    }}