输入: numbers = [2, 7, 11, 15], target = 9
输出: [1,2]
解释: 2 与 7 之和等于目标数 9 。因此 index1 = 1, index2 = 2 。

class Solution {public int searchInsert(int[] nums, int target) {if (nums.length == 0) return -1;int left = 0;//这里用right = nums.length，下面的while循环条件就用left < right//这里如果是right = nums.length - 1，下面的循环条件就是left <= right//这是因为使用right = nums.length循环判等的话会出现越界int right = nums.length;while (left < right) {int mid = left + (right - left) / 2;    //防止left和right太大导致相加除二溢出if (nums[mid] == target) {  //如果target就是nums[mid]的值的话就直接返回return mid;}if (target > nums[mid]) {   //如果target比nums[mid]大，我们向右子区间收缩，所以这里需要left = mid + 1left = mid + 1;}if (target < nums[mid]) {//因为我们上面的right = nums.length，并且循环条件为left < right//所以这里不能使用right = mid - 1，具体的大家可以debug观察结果right = mid;}}return left;}
}

public class Solution {public int[] twoSum(int[] numbers, int target) {int right = numbers.length - 1;//固定一个值，在剩下的元素中找一个数是否等于target - numbers[i]的值for (int i = 0; i < numbers.length; i++) {if (numbers[i] + numbers[right] == target) {return new int[]{i + 1, right + 1};}int index2 = binarySearch(numbers, i + 1, right, target - numbers[i]);if (index2 != -1) {return new int[]{i + 1, index2 + 1};}}return new int[]{-1, -1};}//手写binarySearch()方法，当然也可以直接调用Arrays.binarySearch()方法public int binarySearch(int[] nums, int fromIndex, int toIndex, int target) {int left = fromIndex;int right = toIndex - 1;while (left <= right) {int mid = left + (right - left) / 2;if (nums[mid] == target)return mid;else if (nums[mid] > target) {right = mid - 1;} else if (nums[mid] < target) {left = mid + 1;}}return -1;}
}