# 198 House Robber

記得 Edge Case / 切割問題成小問題

LeetCode 果然 DP 題目都想很久...

You are a professional robber planning to rob houses along a street. 
Each house has a certain amount of money stashed, 
the only constraint stopping you from robbing each of them is that 
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.


input: 一個陣列，每個代表那棟房子有的錢
output: 不能連續抓相鄰的值不燃警察會來，請 output 可以拿到最多的錢

Example 1:

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.
Example 2:

Input: [2,7,9,3,1]
Output: 12
Explanation: Rob house 1 (money = 2), rob house 3 (money = 9) and rob house 5 (money = 1).
             Total amount you can rob = 2 + 9 + 1 = 12.

*/
/**
 * @param {number[]} nums
 * @return {number}
 */
var rob = function(nums) {}

Edge Case

[4, 1, 1, 4]

如何解

本來以為不是奇數就是偶數，結果忽略其他狀況(例如 Edge Case)。這題限制是不能抓隔壁值，所以是兩兩比較，需要紀錄現在為止加起來最大值跟前一個的最大值

let preMax = 0;
let curMax = 0;
let tmpMax = 0;

var rob = function(nums) {
    if(nums.length == 0) return 0;
    if(nums.length == 1) return nums[0]
    let preMax = 0;
    let curMax = 0;
    let tmpMax = 0;
    for(let i = 0;i < nums.length; i++){
        tmpMax = curMax;
        curMax = Math.max(tmpMax, nums[i] + preMax);
        preMax = tmpMax;
    }
    return curMax;
};


console.log(rob([4, 1 ,1,4]))
// faster than 72.97% of JavaScript online submission

學到什麼?

Edge Case 跟切割問題