leetCode 198. House Robber | 動態(tài)規(guī)劃

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.

解題思路：

房間一共有N個，先判斷到目前為止，前i個房間能獲得最多的金錢。

典型的動態(tài)規(guī)劃。

其中轉(zhuǎn)移方程如下：

maxV[i] = max( maxV[i - 2] + a[i],maxV[i-1]);

其中數(shù)組a[i]為第i個房間隱藏的金錢。maxV[i]表示前i個房間能獲得的最多的錢。

代碼如下：

class Solution {
public:
    int rob(vector<int>& nums) 
    {
        //處理特殊情況
    	if (nums.empty())
    		return 0;
    	if (nums.size() == 1)
    		return nums[0];
    	if (nums.size() == 2)
    		return nums[0] > nums[1] ? nums[0] : nums[1];
    	//處理正常情況	
    	int * maxV = new int[nums.size()];
    
    	maxV[0] = nums[0];
    	maxV[1] = nums[0] > nums[1] ? nums[0] : nums[1];
    
    	for (int i = 2 ; i < nums.size() ; ++i)
    	{
    		maxV[i] = max(maxV[i - 2] + nums[i], maxV[i - 1]);
    	}
    
    	int result = maxV[nums.size() - 1];
    	delete maxV;
    	maxV = NULL;
    	return result;
    }
};

2016-08-31 21:49:51