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;
}