(Java) LeetCode 213. House Robber II —— 打家劫舍 II

You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed. All houses at this place are arranged in a circle. That means the first house is the neighbor of the last one. Meanwhile, adjacent houses have security system connected and it will automatically contact the police if two adjacent houses were broken into on the same night.

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.

Example 1:

Input: [2,3,2]
Output: 3
Explanation: You cannot rob house 1 (money = 2) and then rob house 3 (money = 2),
             because they are adjacent houses.

Example 2:

Input: [1,2,3,1]
Output: 4
Explanation: Rob house 1 (money = 1) and then rob house 3 (money = 3).
             Total amount you can rob = 1 + 3 = 4.

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LeetCode 198. House Robber —— 打家劫舍的升级版，但细想想完全一样。首尾不能同时打劫那就不要同时打劫就好。意思是要么从第一家开始劫到倒数第二家，要么从第二家开始劫到最后一家。相当于把198题做了两遍。

 

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Java

class Solution {
    public int rob(int[] nums) {
        if (nums == null || nums.length == 0) return 0;
        if (nums.length == 1) return nums[0];
        int dp1 = nums[0], dp2 = Math.max(dp1, nums[1]);
        int dpS = dp2;
        if (nums.length == 2) return dp2;
        int dp3 = nums[1], dp4 = Math.max(dp3, nums[2]);
        int dpE = dp4;
        for (int i = 2; i < nums.length; i++) {
            if (i != nums.length - 1) {
                dpS = Math.max(dp1 + nums[i], dp2);
                dp1 = dp2;
                dp2 = dpS;
            }     
            if (i != 2) {
                dpE = Math.max(dp3 + nums[i], dp4);
                dp3 = dp4;
                dp4 = dpE;
            }
        }
        return Math.max(dpS, dpE);
    }
}

 

转载于:https://www.cnblogs.com/tengdai/p/9282408.html