198. House Robber
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 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.
这是一道动态分配问题
主要借鉴https://blog.csdn.net/u014593748/article/details/70207883
方法一：
动态规划最重要的是定义状态及状态转移方程，算法的优劣也在此体现。
令dp[i]表示在选择nums[i]的情况下，所能抢到的最大钱数。
则，dp[0]=nums[0]，dp[1]=nums[1]，dp[2]=dp[0]+nums[2]dp[0]=nums[0]，dp[1]=nums[1]，dp[2]=dp[0]+nums[2]
dp[i]=max{dp[i−2]+nums[i],dp[i−3]+nums[i],…,dp[0]+nums[i]}dp[i]=max{dp[i−2]+nums[i],dp[i−3]+nums[i],…,dp[0]+nums[i]}
由于dp[i]表示已经抢了nums[i]，故之前需要从dp[i-2]开始。
会超时
class Solution{
public int rob(int[] num) {
return dynamic(num,num.length-1);
}
public int dynamic(int[] num,int index) {
if(index<0)
return 0;
return Math.max(num[index]+dynamic(num,index-2), dynamic(num,index-1));
}
}
方法二：
class Solution{
public int rob(int[] num) {
int[][] dp = new int[num.length + 1][2];
for (int i = 1; i <= num.length; i++) {
dp[i][0] = Math.max(dp[i - 1][0], dp[i - 1][1]);
dp[i][1] = num[i - 1] + dp[i - 1][0];
}
return Math.max(dp[num.length][0], dp[num.length][1]);
}
}