167. 两数之和 II - 输入有序数组

秒速五厘米 2022-09-10 09:23 115阅读 0赞

[https://leetcode-cn.com/problems/two-sum-ii-input-array-is-sorted/][https_leetcode-cn.com_problems_two-sum-ii-input-array-is-sorted]

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给定一个已按照*** *非递减顺序排列  **的整数数组 `numbers` ，请你从数组中找出两个数满足相加之和等于目标数 `target` 。

函数应该以长度为 `2` 的整数数组的形式返回这两个数的下标值*。*`numbers` 的下标 **从 1 开始计数** ，所以答案数组应当满足 `1 <= answer[0] < answer[1] <= numbers.length` 。

你可以假设每个输入 **只对应唯一的答案** ，而且你 **不可以** 重复使用相同的元素。

**示例 1：**

输入：numbers = [2,7,11,15], target = 9
    输出：[1,2]
    解释：2 与 7 之和等于目标数 9 。因此 index1 = 1, index2 = 2 。

**示例 2：**

输入：numbers = [2,3,4], target = 6
    输出：[1,3]

**示例 3：**

输入：numbers = [-1,0], target = -1
    输出：[1,2]

**提示：**

*  `2 <= numbers.length <= 3 * 104`
 *  `-1000 <= numbers[i] <= 1000`
 *  `numbers` 按 **非递减顺序** 排列
 *  `-1000 <= target <= 1000`
 *  **仅存在一个有效答案**

通过次数278,033提交次数474,041

class Solution {
        public int[] twoSum(int[] numbers, int target) {
            int [] ans = new int [2];
            int start = 0;
            int end = numbers.length-1;
            while(start<end)
            {
                // System.out.println(numbers[start]+" "+numbers[end]);
                if(numbers[start]+numbers[end] == target)
                {
                    ans[0] = start+1;
                    ans[1] = end+1;
                    return ans;
                } 
                else if(numbers[start]+numbers[end]>target) end--;
                else start++;
            }
            return ans;
        }
    }

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[https_leetcode-cn.com_problems_two-sum-ii-input-array-is-sorted]: https://leetcode-cn.com/problems/two-sum-ii-input-array-is-sorted/
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