leetcode 213. House Robber II 抢劫房子II---------- java

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.

 

 

在之前一道题上的改进，就是说首位也是相连的，然后求出最大抢劫金钱。

方法就是在之前方法的基础上，如果最后

per + nums[i] > curr && nums[0] + nums[2] > nums[1]

判断一下。其次，就是分两个数组进行判断。

public class Solution {
    public int rob(int[] nums) {
        if (nums.length == 0){
            return 0;
        }
        if (nums.length == 1){
            return nums[0];
        }
        int[] nums2 = new int[nums.length];
        for (int i = 0; i < nums.length - 1; i++){
            nums2[i] = nums[i + 1];
        }
        nums2[nums.length - 1] = nums[0];
        return Math.max(robber(nums), robber(nums2));
    }
    public int robber(int[] nums){
        int per = nums[0];
        int curr = nums[1];
        for (int i = 2; i < nums.length; i++){
            int num = Math.max(curr, per + nums[i]);
            if (i == nums.length - 1){
                if (per + nums[i] > curr && nums[0] + nums[2] > nums[1]){
                    return Math.max(Math.max(nums[0], nums[i]) + per - nums[0], curr);
                }
            }
            per = curr;
            curr = num;
            
        }
        
        return curr;
    }
}

 

 

可以稍微优化一下。

public class Solution {
    public int rob(int[] nums) {
        if (nums.length == 0){
            return 0;
        }
        if (nums.length == 1){
            return nums[0];
        }
        return Math.max(robber(nums, 0, nums.length - 1), robber(nums, 1, nums.length));
    }
    public int robber(int[] nums,int start, int end){
        int per = 0, curr = 0;
        for (int i = start; i < end; i++){
            int num = Math.max(curr, per + nums[i]);
            per = curr;
            curr = num;
        }
        return curr;
    }
}

 

 

也可以一次循环直接解决。记录第一个点究竟是否选中。

public class Solution {
        public int rob(int[] nums) {
            if(nums.length == 0) return 0;
            if(nums.length == 1) return nums[0];
            int firstInc = nums[0];
            int firstExc = 0;
            int nonFirstInc = 0;
            int nonFirstExc = 0;
            
            for(int i = 1; i < nums.length; i++) {
                int preFirstInc = firstInc;
                firstInc = firstExc + nums[i];
                firstExc = Math.max(preFirstInc, firstExc);
                
                int preNFinc = nonFirstInc;
                nonFirstInc = nonFirstExc + nums[i];
                nonFirstExc = Math.max(preNFinc, nonFirstExc);
            }
            
            int maxInc = Math.min(firstInc, nonFirstInc);
            
            return Math.max(maxInc, firstExc);
        }
    }