167. Two Sum II

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. Please note that 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.

Input: numbers={2, 7, 11, 15}, target=9

Output: index1=1, index2=2

class Solution {    public int[] twoSum(int[] numbers, int target) {        int[] result = new int[2];        HashMap<Integer,Integer> map=new HashMap<Integer,Integer>();            for(int i=0;i<numbers.length;i++){                if(map.containsKey(target-numbers[i])){                        result[0]=i+1;                        result[1]=map.get(target-numbers[i])+1;                        if(result[0]>result[1]){                            int a=result[0];                            result[0]=result[1];                            result[1]=a;                        }                        return result;                }                map.put(numbers[i],i);            }        return result;    }}

采用了hashmap，跟Two Sum那题只是加了一个比较大小的代码。

看到别人的解法，没有用到hashmap，而是用了两个pointer，可以借鉴。毕竟nums这个array已经排好了顺序，可以不用hashmap

public int[] twoSum(int[] numbers, int target) {    int[] indice = new int[2];    if (numbers == null || numbers.length < 2) return indice;    int left = 0, right = numbers.length - 1;    while (left < right) {        int v = numbers[left] + numbers[right];//改成long以免溢出？？        if (v == target) {            indice[0] = left + 1;            indice[1] = right + 1;            break;        } else if (v > target) {            right --;        } else {            left ++;        }    }    return indice;}