First Position of Target

For a given sorted array (ascending order) and a target number, find the first index of this number in O(log n) time complexity.

If the target number does not exist in the array, return -1.

Have you met this question in a real interview?  Yes Example

If the array is [1, 2, 3, 3, 4, 5, 10], for given target 3, return 2.

这是一道二分搜索的题。跟一般的二分搜索的区别在于对于target == nums[mid]的情况，不会直接返回，而是继续执行，直到low，high重合。如果targe == nums[mid]， 那么为了找下边界，所以要使high变低，low不变，这样能进一步缩小，从而找到目标值的下边界。

代码：

public int binarySearch(int[] nums, int target) { //write your code here if(nums == null || nums.length == 0) return -1; int low = 0; int high = nums.length-1; while(low<high){ int mid = low +(high - low)/2; if(nums[mid] > target){ high = mid -1; }else if(nums[mid] < target){ low = mid +1; }else{ high = mid; } } return nums[low] == target?low:-1; }