Binary search (3) -- 4Sum II, Kth Smallest Element in a BST

4Sum II

Given four lists A, B, C, D of integer values, compute how many tuples (i, j, k, l) there are such that A[i] + B[j] + C[k] + D[l] is zero.

To make problem a bit easier, all A, B, C, D have same length of N where 0 ≤ N ≤ 500. All integers are in the range of -228 to 228 - 1 and the result is guaranteed to be at most 231 - 1.

Example:

Input:A = [ 1, 2]B = [-2,-1]C = [-1, 2]D = [ 0, 2]Output:2

1. 分别计算A,B和C,D的所有可能的和，然后遍历所有可能性。

2. 开始想能不能用2Sum的思想去做。将C和D先进行sort。但是2Sum那道题本身就是排序的，在这道题中进行排序会消耗很多时间。

3. 非常直接的思想是O(n^4)，但是数组加和会有重复结果。可以从这个角度去优化

4. 但是就算数组加和没有重复结果。O(n^4)也将数组加和的结果重复计算了很多次，事实上我们并不需要这样。

    int fourSumCount(vector<int>& A, vector<int>& B, vector<int>& C, vector<int>& D) {        int rst = 0;        unordered_map<int, int> map1;        for (int a : A){            for (int b : B){                int sum = a + b;                if (map1.find(sum) != map1.end()) map1[sum]++;                else map1[sum] = 1;            }        }        for (int c : C){            for (int d : D){                int sum = c + d;                int target = (-1) * sum;                if (map1.find(target) != map1.end())                    rst += map1[target];            }        }        return rst;    }

Kth Smallest Element in a BST

解法1：中序遍历。即从小到大遍历整个查找树

    bool traverse(TreeNode* root, int k, int& rst, int& count){        if (root == NULL) return false;        bool found = traverse(root->left, k, rst, count);        count++;        if (count == k){   //找到第k小            rst = root->val;            return true;        }         if(!found) found = traverse(root->right, k, rst, count);  //如果已经找到，就无需搜索另一边        return found;    }        int kthSmallest(TreeNode* root, int k) {        int rst = 0;        int count = 0;        traverse(root, k, rst, count);        return rst;    }

解法2：快速选择的思想。但是countNodes会有重复计算。使用hashmap来避免重复计算

    int kthSmallest(TreeNode* root, int k) {        unordered_map<int,int> small;        return kth(root, k, small);    }        int kth(TreeNode* root, int k, unordered_map<int, int>& small){        int count = countNodes(root->left, small);        if (k <= count) {            return kth(root->left, k, small);        } else if (k > count + 1) {            return kth(root->right, k-1-count, small); // 1 is counted as current node        }        return root->val;    }        int countNodes(TreeNode* root, unordered_map<int, int>& small) {        if (root == NULL) return 0;        if (small.find(root->val) != small.end())             return small[root->val];        int count = 1 + countNodes(root->left, small) + countNodes(root->right, small);        small[root->val] = count;        return count;    }