LeetCode //C - 1. Two Sum

1. Two Sum

Given an array of integers nums and an integer target, return indices of the two numbers such that they add up to target.

You may assume that each input would have exactly one solution, and you may not use the same element twice.

You can return the answer in any order.

Example 1:

Input: nums = [2,7,11,15], target = 9
Output: [0,1]
Explanation: Because nums[0] + nums[1] == 9, we return [0, 1].

Example 2:

Input: nums = [3,2,4], target = 6
Output: [1,2]

Example 3:

Input: nums = [3,3], target = 6
Output: [0,1]

Constraints:

2 < = n u m s . l e n g t h < = 1 0 4 2 <= nums.length <= 10^4 2<=nums.length<=104
− 1 0 9 < = n u m s [ i ] < = 1 0 9 -10^9 <= nums[i] <= 10^9 −109<=nums[i]<=109
− 1 0 9 < = t a r g e t < = 1 0 9 -10^9 <= target <= 10^9 −109<=target<=109
Only one valid answer exists.

From: LeetCode

Link: 1. Two Sum

Solution:

Ideas：

In this implementation, we loop through each element and then loop through the rest of the elements to find a pair that sums up to the target. Once the pair is found, we store their indices in the result array and return it.

Code:

c 复制代码

/**
 * Note: The returned array must be malloced, assume caller calls free().
 */
int* twoSum(int* nums, int numsSize, int target, int* returnSize) {
    int* result = malloc(2 * sizeof(int));  // Allocate memory for the result
    *returnSize = 2;  // Set the return size to 2

    for (int i = 0; i < numsSize - 1; i++) {
        for (int j = i + 1; j < numsSize; j++) {
            if (nums[i] + nums[j] == target) {
                result[0] = i;
                result[1] = j;
                return result;
            }
        }
    }

    // In case no solution is found, though the problem statement guarantees one solution
    result[0] = -1;
    result[1] = -1;
    return result;
}