198. 打家劫舍 两种解法

题目描述

 

dp[0]=nums[0]

dp[1]=max(nums[0],nums[1])​

dp[i]=max(dp[i−2]+nums[i],dp[i−1])

其实这种题目的难度就在于题目描述的公式

 

解法1：利用辅助数组

#define MAX(x, y) ((x)>(y)?(x):(y))
class Solution {
public:
    int rob(vector& nums) {
        int len = nums.size();
        if (len == 0)
            return 0;
        if (len == 1)
            return nums[0];
        if (len == 2)
            return MAX(nums[0], nums[1]);
        
        vector dp(len, 0);
        dp[0] = nums[0];
        dp[1] = MAX(nums[0], nums[1]);
        for (int i = 2; i < len; i++)
            dp[i] = MAX(dp[i-2] + nums[i], dp[i-1]);

        return dp[len-1];
    }
};

解法2：利用两个变量记录历史结果

#define MAX(x, y) ((x)>(y)?(x):(y))
class Solution {
public:
    int rob(vector& nums) {
        int len = nums.size();
        if (len == 0)
            return 0;
        if (len == 1)
            return nums[0];
		if (len == 2)
            return MAX(nums[0], nums[1]);
        
        int one = nums[0];
        int two = MAX(one, nums[1]);        //初始化
        int max = two;
        for (int i = 2; i < len; i++) {
            max = MAX(one + nums[i], two);
            one = two;
            two = max;
        }

        return max;
    }
};