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 = 9Output: [1,2]Explanation: The sum of 2 and 7 is 9. Therefore index1 = 1, index2 = 2.

/** * 简单一点 我们就全部遍历一遍 * @param numbers * @param target * @return */public static int[] twoSum(int[] numbers, int target) {    int sum[] = new int[2];    int i= 0,j=0;    //这里使用break加标志符号跳出指定循环，其实也可以加一个boolean值判断    out:for (;i
    
      target){                //j=numbers.length-1;                break;            }            if(numbers[i]+numbers[j] == target){                break out;            }        }    }    sum[0]=++i;    sum[1]=++j;    return sum;}/*  这样的时间复杂度就降低了很多 O（n），比起上面的O（n^2） */public static int[] twoSum3(int[] numbers, int target) {    int low = 0;    int high = numbers.length - 1;    while(low < high){        if(numbers[low] + numbers[high] < target){            low++;            continue;        }        else if (numbers[low] + numbers[high] > target){            high--;            continue;        }        else {            return new int[] {low+1, high+1};        }    }    return null;}
    

git：https://github.com/woshiyexinjie/leetcode-xin/tree/master/src/main/java/com/helloxin/leetcode/algorithms