Given an array of integers nums which is sorted in ascending order, and an integer target, write a function to search target in nums. If target exists, then return its index. Otherwise, return -1.

You must write an algorithm with O(log n) runtime complexity.

Example 1:

Input: nums = [-1,0,3,5,9,12], target = 9 Output: 4 Explanation: 9 exists in nums and its index is 4 Example 2:

Input: nums = [-1,0,3,5,9,12], target = 2 Output: -1 Explanation: 2 does not exist in nums so return -1

Constraints:

1 <= nums.length <= 104 -104 < nums[i], target < 104 All the integers in nums are unique. nums is sorted in ascending order.

来源：力扣（LeetCode） 链接：https://leetcode-cn.com/problems/binary-search 著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

class Solution {
    public int search(int[] nums, int target) {
        int head=0;
        int tail=nums.length-1;
        while(true){
            if(head>tail){
                return -1;
            }
            int mid=(head+tail)/2;
            if(target==nums[mid]){
                return mid;
            }else if(target<nums[mid]){
                tail=mid-1;
            }else{
                head=mid+1;
            }
        }
    }
}

Given an array of integers nums which is sorted in ascending order, and an integer target, write a function to search target in nums. If target exists, then return its index. Otherwise, return -1.

You must write an algorithm with O(log n) runtime complexity.

Example 1:

Input: nums = [-1,0,3,5,9,12], target = 9 Output: 4 Explanation: 9 exists in nums and its index is 4 Example 2:

Input: nums = [-1,0,3,5,9,12], target = 2 Output: -1 Explanation: 2 does not exist in nums so return -1

Constraints:

1 <= nums.length <= 104 -104 < nums[i], target < 104 All the integers in nums are unique. nums is sorted in ascending order.

来源：力扣（LeetCode） 链接：https://leetcode-cn.com/problems/binary-search 著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

class Solution {
    public int search(int[] nums, int target) {
        int head=0;
        int tail=nums.length-1;
        while(true){
            if(head>tail){
                return -1;
            }
            int mid=(head+tail)/2;
            if(target==nums[mid]){
                return mid;
            }else if(target<nums[mid]){
                tail=mid-1;
            }else{
                head=mid+1;
            }
        }
    }
}