[lintcode 14] 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：
题中给出一个有序数组，每个元素不一定唯一，找到第一次出现target的位置。
这道题处理的关键点就是当nums[mid] == target的时候，不是将mid返回，
而是把end移动的mid点，继续向前查找看有无相同的元素。

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