【LEETCODE】38、167题，Two Sum II - Input array is sorted

package y2019.Algorithm.array;
 
/**
 * @ProjectName: cutter-point
 * @Package: y2019.Algorithm.array
 * @ClassName: TwoSum2
 * @Author: xiaof
 * @Description: 167. Two Sum II - Input array is sorted
 * 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.
 *
 * 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.
 *
 * @Date: 2019/7/2 10:22
 * @Version: 1.0
 */
public class TwoSum2 {
 
    public int[] solution(int[] numbers, int target) {
        //求两数之和就是对应的target，并且反馈对应的下标
        int index1 = 0, index2 = 0;
        //这里其实也是遍历，但是已知target，那么我们每次只要获取到第一个数据的下表，然后对后面的数据进行二分查找判断是否有对应的数据就可以了
        for(int i = 0; i < numbers.length; ++i) {
            index1 = i;
            int num1 = numbers[i];
            //二分查找目标
            int targetNum = target - num1;
            int start = i + 1, end = numbers.length - 1;
            while(start < end) {
                int mid = start + ((end - start) >>> 1);
                if(numbers[mid] == targetNum) {
                    index2 = mid;
                    break;
                } else if (numbers[mid] > targetNum) {
                    //如果目标位置比目标数据大，说明要找的数据在前面
                    end = mid;
                } else {
                    //如果比目标数据小，说明再后面，然后我们mid的位置已经做了比较，所以后移一位
                    //因为mid = start + (end - start) >>> 1; 可能正好等于start
                    start = mid + 1;
                }
            }
 
            if(start == end) {
                //如果首尾相等，那么手动判断一下
                if(targetNum == numbers[start]) {
                    index2 = start;
                }
            }
 
            if(index2 != 0) {
                break;
            }
        }
 
        int result[] = new int[2];
        result[0] = index1 + 1;
        result[1] = index2 + 1;
        return result;
    }
 
    //网上大牛最快，也是占内存最小的算法,有点类似快排的思想
    public int[] twoSum(int[] numbers, int target) {
        int []res = new int [2];
        int index1 = 0;
        int index2 = numbers.length - 1;
        while (index2 > index1){
            if (numbers[index1] + numbers[index2] > target){
                index2 --;
            }
            else if(numbers[index1] + numbers[index2] < target){
                index1 ++;
            }else{
                res[0] = index1+1;
                res[1] = index2+1;
                break;
            }
        }
        return res;
    }
 
    public static void main(String args[]) {
 
        int pres[] = {5,25,75};
        int target = 100;
        System.out.println(new TwoSum2().solution(pres, target));
 
    }
 
}