House RobberII; DP;

In this problem, house are arranged in a circle, robber should not invade into two adjacent houses. Compared to the former problem, we need to consider the problem that the optimal option include the first one and the last house, which is not allowede in this problem.Thus, we need divide the option to two sub-options, one is an optimal option exclude the last house, the other is an option that excludes the first house. By compare the value of the two option, we can get the optimal solution for this problem.

Code:

public class Solution {
    public int rob(int[] nums) {
        int len = nums.length;
        if(len == 0) return 0;
        if(len == 1) return nums[0];
        return Math.max(robOption(nums, 0, len-2), robOption(nums, 1, len-1)); 
    }
    
    public int robOption(int[] nums, int start, int end){
        int prev2 = 0, prev = 0;
        for(int i = start; i <= end; i++){
            int temp = nums[i] + prev2;
            if(temp > prev) {
                prev2 = prev;
                prev = temp;
            }
            else{
                prev2 = prev;
            }
        }
        return prev;
    }
}

　　

转载于:https://www.cnblogs.com/5683yue/p/5226275.html