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].文章來源：http://www.zghlxwxcb.cn/news/detail-612646.html
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
文章來源地址http://www.zghlxwxcb.cn/news/detail-612646.html
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:
/**
* 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;
}
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