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 

大意就是求数组中不相邻元素的最大和。

运用动态规划思想，数组从前向后遍历，nums[i]代表到当前为止的最大和，显然nums[0]就是本身的值，到nums[1]时，可以选择不偷，那么收益就是nums[0],如果偷，那么就选择不偷前面一个，受益就是nums[1],所以nums[1]的最大收益就是Math.max(nums[0], nums[1])。

当i>2时， 如果不偷i，那么收益就是nums[i-1]

如果偷了i，那么就不能偷i-1，收益就是nums[i-2]+nums[i],

得出最后的状态方程：nums[i]=Math.max(nums[i-1], nums[i-2]+nums[i])

public int rob(int[] nums) {
if(nums.length<2){
return nums.length==0? 0:nums[0];
}
nums[1]=Math.max(nums[0], nums[1]);
for(int i=2; i<nums.length; i++){
nums[i]=Math.max(nums[i-1], nums[i-2]+nums[i]);
}
return nums[nums.length-1];
}

总结：刚开始做动态规划的题，还不太熟悉，最重要的就是找出递推关系。