LeetCode //167. Two Sum II - Input Array Is Sorted

167. Two Sum II - Input Array Is Sorted

Given a 1-indexed array of integers numbers that is already sorted in non-decreasing order, find two numbers such that they add up to a specific target number. Let these two numbers be numbers[index1] and numbers[index2] where 1 <= index1 < index2 < numbers.length.

Return the indices of the two numbers, index1 and index2 , added by one as an integer array [index1, index2] of length 2.

The tests are generated such that there is exactly one solution. You may not use the same element twice.

Your solution must use only constant extra space.

Example 1:

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. We return [1, 2].

Example 2:

Input: numbers = [2,3,4], target = 6
Output: [1,3]
Explanation: The sum of 2 and 4 is 6. Therefore index1 = 1, index2 = 3. We return [1, 3].

Example 3:

Input: numbers = [-1,0], target = -1
Output: [1,2]
Explanation: The sum of -1 and 0 is -1. Therefore index1 = 1, index2 = 2. We return [1, 2].

Constraints:

2 < = n u m b e r s . l e n g t h < = 3 ∗ 1 0 4 2 <= numbers.length <= 3 * 10^4 2<=numbers.length<=3∗104
-1000 <= numbers[i] <= 1000
numbers is sorted in non-decreasing order.
-1000 <= target <= 1000
The tests are generated such that there is exactly one solution.

From: LeetCode

Link: 167. Two Sum II - Input Array Is Sorted

Solution:

Ideas：

We start with one pointer at the beginning of the array and another at the end. If the sum of the numbers at the two pointers is less than the target, we move the pointer at the beginning forward to increase the sum. If the sum is greater than the target, we move the pointer at the end backward to decrease the sum. We continue this process until we find two numbers that add up to the target.

Code:

c 复制代码

/**
 * Note: The returned array must be malloced, assume caller calls free().
 */
int* twoSum(int* numbers, int numbersSize, int target, int* returnSize){
    *returnSize = 2;
    int* result = (int*) malloc(*returnSize * sizeof(int));
    int left = 0, right = numbersSize - 1;

    while(left < right){
        int sum = numbers[left] + numbers[right];

        if(sum == target){
            result[0] = left + 1;  // +1 because the problem is 1-indexed
            result[1] = right + 1; // +1 because the problem is 1-indexed
            return result;
        }
        else if(sum < target)
            left++;
        else
            right--;
    }

    // This point should never be reached because the problem guarantees a solution
    return NULL;
}