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

.

Example

If the array is

[1, 2, 3, 3, 4, 5, 10]

, for given target

3

, return

2

.

分析：
很明显的用binary search, 但是因为要找第一个，所以，我们需要判断一下。

class Solution {
/**
* @param nums: The integer array.
* @param target: Target to find.
* @return: The first position of target. Position starts from 0.
*/
public int binarySearch(int[] nums, int target) {
if (nums == null || nums.length == 0) return -1;

int start = 0;
int end = nums.length - 1;

while (start <= end) {
int mid = start + (end - start) / 2;
if (nums[mid] == target) {
if (mid != 0 && nums[mid - 1] == target) {
end = mid - 1;
} else {
return mid;
}
} else if (nums[mid] < target) {
start = mid + 1;
} else {
end = mid - 1;
}
}
return -1;
}
}

参考请注明出处： cnblogs.com/beiyeqingteng/