167. Two Sum II - Input Array Is Sorted 两数之和 II - 输入有序数组

题目地址：https://leetcode.com/problems/two-sum-ii-input-array-is-sorted/#/descriptionopen in new window

题目描述

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.

Thefunction 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, index2 = 2.

题目大意

在已经排好序的数组中，找到两个数字的和等于target，返回两个数字的基于1开始的索引。

解题方法

Java解法

已经排序好了的数组，找出目标和。可以挨个试，时间复杂度是O(n^2)，复杂度太高。可以用双指针，同时向中间靠拢，肯定能达到目标的和。

public class Solution {
    public int[] twoSum(int[] numbers, int target) {
        int index[] = new int[2];
        int left = 0;
        int right = numbers.length - 1;
        while(left < right){
            long temp = numbers[left] + numbers[right];
            if(temp == target){
                index[0] = left + 1;
                index[1] = right + 1;
                break;
            }else if(temp < target){
                left++;
            }else{
                right--;
            }
        }
        return index;
    }
}

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Python解法

二刷，python解法。

class Solution:
    def twoSum(self, numbers, target):
        """
        :type numbers: List[int]
        :type target: int
        :rtype: List[int]
        """
        N = len(numbers)
        left, right = 0, N - 1
        while left < right:
            cursum = numbers[left] + numbers[right]
            if cursum == target:
                return [left + 1, right + 1]
            elif cursum < target:
                left += 1
            else:
                right -= 1
        return [0, 0]

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