[LeetCode][JavaScript]House Robber II

House Robber II

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.

https://leetcode.com/problems/house-robber-ii/

 

 

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和House Robber新相比，房子首尾相连。

动态规划和上一题一样的思路：http://www.cnblogs.com/Liok3187/p/4842444.html

分成2次，第一次从第一间房子开始到倒数第二间结束；

第二次从第二间开始到最后一间结束。

粗暴地写两遍：

/**
 * @param {number[]} nums
 * @return {number}
 */
var rob = function(nums) {
    if(nums.length === 0){
        return 0;
    }else if(nums.length === 1){
        return nums[0];
    }
    var dp = [];
    dp[0] = 0;
    dp[1] = nums[0];
    for(var i = 2; i <= nums.length - 1; i++){
        dp[i] = Math.max(dp[i - 2] + nums[i - 1], dp[i - 1]);
    }
    var candidate = dp[nums.length - 1];

    dp = [];
    dp[1] = 0;
    dp[2] = nums[1];
    for(var i = 3; i <= nums.length; i++){
        dp[i] = Math.max(dp[i - 2] + nums[i - 1], dp[i - 1]);
    }

    return Math.max(candidate, dp[nums.length]);
};

 

封装成方法：

/**
 * @param {number[]} nums
 * @return {number}
 */
var rob = function(nums) {
    if(nums.length === 0){
        return 0;
    }else if(nums.length === 1){
        return nums[0];
    }
    return Math.max(robbing(1, nums.length - 1), robbing(2, nums.length));

    function robbing(start, end){
        var dp = [];
        dp[start - 1] = 0;
        dp[start] = nums[start - 1];
        for(var i = start + 1; i <= end; i++){
            dp[i] = Math.max(dp[i - 2] + nums[i - 1], dp[i - 1]);
        }
        return dp[end];
    }   
};