leetcode-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

中文大意:

给定一个已按升序排序的整数数组，找到两个数字，使它们相加得到一个特定的目标数字。函数two和应该返回两个数字的指标，使它们相加得到目标值，其中index1必须小于index2。

注记:您返回的答案(index1和index2)不是基于零的。您可以假设每个输入都只有一个解决方案，并且可能不会两次使用相同的元素。

解题思路:

一个数组，取两个值相加得到target,由于这个数组它的顺序是有序的，所以分别去前面的值与后面的值进行相加，当得到result结果时，我们可以返回一个result数组值，即返回最后的结果

代码如下

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class Solution {     public int[] twoSum(int[] numbers, int target) {         // 最后的返回的数组，记录结果数组由那两个位置的值进行相加的到         int[] result = new int[2];         // 前向的标记         int count1 = 0;         // 后向的标记         int count2 = numbers.length - 1;         // 依次遍历数组         while (count1 < count2) {             // 得到result时，返回两个结果位置             if (numbers[count1] + numbers[count2] == target) {                 result[0] = count1 + 1;                 result[1] = count2 + 1;                 break;             }             // 结果大于target时，将后面的count2向前移             else if (numbers[count1] + numbers[count2] > target) {                 count2--;             }             // 结果小于target时，将前面的count1向后移             else {                 count1++;             }         }         return result;     } }