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.
Example：
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.
解釋下題目：
給定一個數組代表每個房子的金額，不能同時搶連續的兩個房子，求最大。
1. DP
實際耗時：0ms
public int rob(int[] nums) {
if (nums == null || nums.length == 0) {
return 0;
}
if (nums.length == 1) {
return nums[0];
}
int[][] dp = new int[nums.length][2];
//init
dp[0][0] = 0;
dp[0][1] = nums[0];
for (int i = 1; i < nums.length; i++) {
// if robber do not rob the house
dp[i][0] = Math.max(dp[i - 1][0], dp[i - 1][1]);
// robber rob the house
dp[i][1] = dp[i - 1][0] + nums[i];
}
return Math.max(dp[nums.length - 1][0], dp[nums.length - 1][1]);
}
踩過的坑：[]
??思路很簡單，創建一個二維數組，dp[i][0]代表不搶第i個房間的，而dp[i][1]則代表搶這個房間，不停繼續即可。做這種題目，包括類似遞歸的解法，一定要假設自己已經解出來了，然后再做，否則特別容易進入死胡同。
