leetcode 198House Robber（简单动态规划解法）

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.

代码如下：

public class Solution {
    public int rob(int[] nums) {  
    if(nums.length==0) return 0;
    if(nums.length==1) return nums[0];

    //Initialize an arrays to store the money
	int[] mark = new int[nums.length];

    //We can infer the formula from problem:mark[i]=max(num[i]+mark[i-2],mark[i-1])
    //so initialize two nums at first.
	mark[0] = nums[0];
	mark[1] = Math.max(nums[0], nums[1]);

    //Using Dynamic Programming to mark the max money in loop.
	for(int i=2;i

迭代的解法，空间复杂度降为O（1）代码如下：

class Solution {
public:
    int rob(vector& nums) {
        if (nums.size() == 0) return 0;
        if (nums.size() == 1) return nums[0];
        if (nums.size() == 2) return max(nums[0], nums[1]);
        int prev1 = nums[1], prev2 = nums[0], prev1_no_rob = nums[0], cur;
        for (int i = 2; i < nums.size(); i++)
        {
            cur = max(prev1_no_rob + nums[i], max(prev1, prev2 + nums[i]));
            prev1_no_rob = max(prev2, prev1);
            prev2 = prev1;
            prev1 = cur;
        }
        return cur;
    }
};