[LintCode] 14 First Position of Target

Description
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.


Challenge
If the count of numbers is bigger than 2^32, can your code work properly?

思路
二分法。找到nums[mid] == target时，因为我们是要找第一个符合的位置，所以要接着往前找，并包括这个，也就是end = mid。

Code

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) {
        //write your code here
        
        if (nums == null || nums.length == 0) {
            return -1;
        }
        
        int start = 0;
        int 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;
            }
        }
        
        if (nums[start] == target) {
            return start;
        }
        if (nums[end] == target) {
            return end;
        }
        return -1;
    }
}