Search for a Range 解答

Question

Given a sorted array of integers, find the starting and ending position of a given target value.

Your algorithm's runtime complexity must be in the order of O(log n).

If the target is not found in the array, return [-1, -1].

For example,
Given [5, 7, 7, 8, 8, 10] and target value 8,
return [3, 4].

Solution

Use binary search, first, find left position, then, find right position.

public class Solution {
    public int[] searchRange(int[] nums, int target) {
        int[] result = new int[2];
        result[0] = -1;
        result[1] = -1;
        if (nums == null || nums.length < 1)
            return result;
        int start = 0, end = nums.length - 1, mid, first, last;
        // Find first position of target
        while (start + 1 < end) {
            mid = (end - start) / 2 + start;
            if (nums[mid] >= target)
                end = mid;
            else
                start = mid;
        }
        if (nums[start] == target)
            result[0] = start;
        else if (nums[end] == target)
            result[0] = end;
        
        // Find last position of target
        start = 0;
        end = nums.length - 1;
        while (start + 1 < end) {
            mid = (end - start) / 2 + start;
            if (nums[mid] <= target)
                start = mid;
            else
                end = mid;
        }
        if (nums[end] == target)
            result[1] = end;
        else if (nums[start] == target)
            result[1] = start;
        return result;
    }
}