Two Sum & Two Sum II - Input array is sorted & Two Sum IV - Input is a BST

1. Two Sum

Given an array of integers, return indices of the two numbers such that they add up to a specific target.You may assume that each input would have exactlyone solution, and you may not use the same element twice.

Example:

Given nums = [2, 7, 11, 15], target = 9,

Because nums[0] + nums[1] = 2 + 7 = 9, return [0, 1].

 1 int* twoSum(int* nums, int numsSize, int target) {
 2     int *sValue = (int *)malloc(2 * sizeof(int));
 3     for(int i = 1; i < numsSize; i++)
 4     {
 5         for(int j = 0; j < i; j++)
 6         {
 7             if(nums[i] + nums[j] == target)
 8             {
 9                 sValue[0] = i;
10                 sValue[1] = j;
11                 return sValue;
12             }
13         }
14     }
15     return NULL;
16 }

解题思路：

嵌套两层循环：

　　第一层：1 <= i <= numsSize

　　第二层： 0 <= j < (i - 1)因为i的取值是第二个及后面的数据，那么j就要取i前面的数据与i相加才能避免使数据做重复的相加

　　当nums[i] + nums[j] == target 成立就可得到答案

167. Two Sum II - Input array is sorted

Given an array of integers that is already sorted in ascending order, find two numbers such that they add up to a specific target number.

The function twoSum should return indices of the two numbers such that they add up to the target, where index1 must be less than index2.

Note:

Your returned answers (both index1 and index2) are not zero-based.
You may assume that each input would have exactly one solution and you may not use the same element twice.

Example:

Input: numbers = [2,7,11,15], target = 9
Output: [1,2]
Explanation: The sum of 2 and 7 is 9. Therefore index1 = 1, index2 = 2.

int* twoSum(int* numbers, int numbersSize, int target, int* returnSize) {
    *returnSize = 2;
    int *Array = NULL;
    for(int i = 1; i < numbersSize; i++)
    {
        for(int j = 0; j < i; j++)
        {
            if(numbers[i] + numbers[j] == target)
            {
                Array = (int *)malloc(*returnSize * sizeof(int));
                Array[0] = j + 1;
                Array[1] = i + 1;
                return Array;
            }
        }
    }
    return NULL;
}

653. Two Sum IV - Input is a BST

Given a Binary Search Tree and a target number, return true if there exist two elements in the BST such that their sum is equal to the given target.

Example 1:

Input: 
    5
   / \
  3   6
 / \   \
2   4   7

Target = 9

Output: True

Example 2:

Input: 
    5
   / \
  3   6
 / \   \
2   4   7
Target = 28
Output: False

 1 bool findValue(struct TreeNode* root, struct TreeNode* temp, int k)
 2 {
 3     if(!temp)
 4     {
 5         return false;
 6     }
 7     if(temp->val == k && temp != root)
 8     {
 9         return true;
10     }
11     if(k > temp->val)
12     {
13         return findValue(root, temp->right, k);
14     }
15     else
16     {
17         return findValue(root, temp->left, k);
18     }
19 }
20 
21 bool findX(struct TreeNode* root, struct TreeNode* temp, int k)
22 {
23     if(!root)
24     {
25         return false;
26     }
27     if(findValue(root, temp, k - root->val))
28     {
29         return true;
30     }
31     else
32     {
33         return (findX(root->left, temp, k) || findX(root->right, temp, k));
34     }
35 }
36 
37 bool findTarget(struct TreeNode* root, int k) {
38     struct TreeNode* pTemp = root;
39     return findX(root, pTemp, k);
40 }

代码27行的k - root->val表示目标数减去二叉树中某一个数剩下的那个数据，如果递归查找树能找到与k - root->val相等的数并且不是同一个节点的数据（第7行有做判断）说明存在两个相加等于目标的数。