【leetcode】House Robber I && II

/** 
 * @author        johnsondu 
 * @time          2015.8.21 17:00 
 * @description   simple dynamic programming, dp[i] = max(dp[i-1], dp[i-2] + num[i]);
 * @url           https://leetcode.com/problems/house-robber/
 */  

class Solution {
public:
    int max(int a, int b) {
        return a > b ? a : b;
    }
    int rob(vector<int>& nums) {
        int len = nums.size();
        if(len < 1) return 0;
        if(len < 2) return nums[0];
        int *dp = new int[len];
        dp[0] = nums[0];
        dp[1] = max(dp[0], nums[1]);
        for(int i = 2; i < len; i ++) {
            dp[i] = max(dp[i-1], dp[i-2] + nums[i]);
        }
        return dp[len-1];
    }
};

/** 
 * @author        johnsondu 
 * @time          2015.8.21 17:00 
 * @description   simple dynamic programming, plus a circle in it.
 * @url           https://leetcode.com/problems/house-robber-ii/
 */   

class Solution {
public:
    int max(int a, int b) {
        return a > b ? a : b;
    }
    
    int rob(vector<int>& nums) {
        int len = nums.size();
        if(len == 0) return 0;
        if(len == 1) return nums[0];
        // record the maximum result exclude the last element in nums.
        int *dp1 = new int[len + 1];
        // recore the maximum result exclude the first element in nums.
        int *dp2 = new int[len + 1];
        
        memset(dp1, 0, sizeof(dp1));
        memset(dp2, 0, sizeof(dp2));
        dp1[1] = nums[0];
        dp2[2] = nums[1];
        for(int i = 2; i < len; i ++) {
            dp1[i] = max (nums[i-1] + dp1[i-2], dp1[i-1]);
        }
        for(int i = 3; i<= len; i ++) {
            dp2[i] = max(nums[i-1] + dp2[i-2], dp2[i-1]);
        }
        return max(dp1[len-1], dp2[len]);
    }
};