作者: 负雪明烛
id: fuxuemingzhu
个人博客: http://fuxuemingzhu.cn/


题目地址:https://leetcode.com/problems/two-sum-ii-input-array-is-sorted/#/description

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.

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, 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;
}
}

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]

日期

2017 年 4 月 18 日
2018 年 11 月 14 日 —— 又到周五了

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