LeetCode Learning 8

337. House Robber III

The thief has found himself a new place for his thievery again. There is only one entrance to this area, called the "root." Besides the root, each house has one and only one parent house. After a tour, the smart thief realized that "all houses in this place forms a binary tree". It will automatically contact the police if two directly-linked houses were broken into on the same night.

Determine the maximum amount of money the thief can rob tonight without alerting the police.

还是House Robber，这次从数组变成了二叉树，类似的，也是不能选择相邻的2个节点。由于是树状结构，需要使用递归方法，对每个节点保存2个值包含此节点的最大和，以及不包含此节点的最大和res[1]以及res[0]。对root，root的res[1]等于root->left的res[0]+root->right的res[0]+root的值。而res[0]则等于root->left的max(res[0],res[1]) + root->的max(res[0],res[1])。代码如下：

int rob(TreeNode* root) {        vector<int> res = dfs(root);        return max(res[0], res[1]);    }    vector<int> dfs(TreeNode *root) {        if (!root) return vector<int>(2, 0);        vector<int> left = dfs(root->left);        vector<int> right = dfs(root->right);        vector<int> res(2, 0);        res[0] = max(left[0], left[1]) + max(right[0], right[1]);        res[1] = left[0] + right[0] + root->val;        return res;    }

377. Combination Sum IV

Given an integer array with all positive numbers and no duplicates, find the number of possible combinations that add up to a positive integer target.

题意为给出一组正整数与一个目标正整数，求出所有可能的数组中的正整数组合成目标的组合数。依然可以使用动态规划，转移方程中需要加入目标减去数组中每个数的状态。即dp[i]=Σdp[i-nums[k]]。设数组中整数数目为m，目标大小为n，这样的话复杂度会是O(mn)。

最终代码如下：

class Solution {public:    int combinationSum4(vector<int>& nums, int target) {        vector<int> dp(target + 1);        dp[0] = 1;        for (int i = 1; i <= target; ++i) {            for (int j=0;j<nums.size();j++) {                if (i >= nums[j]) dp[i] += dp[i - nums[j]];            }        }        return dp.back();    }};