【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]);
}
};