从已排序数组中求两数和等于输入值TwoSumII167 -leetcode

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.

/**
 * 简单一点 咱们就所有遍历一遍
 * @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，其实也能够加一个boolean值判断
    out:for (;i<numbers.length;i++){
        for(j=i+1;j<numbers.length;j++){
            if(numbers[i]+numbers[j] == target){
                break out;
            }
        }
    }
    sum[0]=++i;
    sum[1]=++j;
    return sum;
}

/**
 * 在1的基础上优化一下
 * @param numbers
 * @param target
 * @return
 */
public static int[] twoSum2(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，其实也能够加一个boolean值判断
    out:for (;i<numbers.length;i++){
        for(j=i+1;j<numbers.length;j++){
            //若是是已经排好序的 那么大于target的时候就不用比较了
            if(numbers[i]+numbers[j] > 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下降了不少 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