[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?

4/26/2017

算法班

不是很明白challenge说数字大于2^32的意思，二分法是合理的吗？

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;
        if (target > nums[nums.length - 1] || target < nums[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 {
                start = mid;  // to find the first position, we need to include target in the search range
            }
        }
        
        if (nums[start] == target) return start;
        else if (nums[end] == target) return end;
        return -1;
    }
}