Problem

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 (JAVA)

public class Solution {
    public int[] searchRange(int[] nums, int target) {
        // 找到左邊界
        int front = search(nums, target, "front");
        // 找到右邊界
        int rear = search(nums, target, "rear");
        int[] res = {front, rear};
        return res;
    }
    
    public int search(int[] nums, int target, String type){
        int min = 0, max = nums.length - 1;
        while(min <= max){
            int mid = min + (max - min) / 2;
            if(nums[mid] > target){
                max = mid - 1;
            } else if(nums[mid] < target){
                min = mid + 1;
            } else {
                // 對于找左邊的情況，要判斷左邊的數(shù)是否重復(fù)
                if(type == "front"){
                    if(mid == 0) return 0;
                    if(nums[mid] != nums[mid - 1]) return mid;
                    max = mid - 1;
                } else {
                // 對于找右邊的情況，要判斷右邊的數(shù)是否重復(fù)
                    if(mid == nums.length - 1) return nums.length - 1;
                    if(nums[mid] != nums[mid + 1]) return mid;
                    min = mid + 1;
                }
            }
        }
        //沒找到該數(shù)返回-1
        return -1;
    }
}

Discussion

這道題的本質(zhì)是就是找到相同值的上下限位置，基本的方法就是二分查找。在找到第一個值之后，我們需要確定這個值的上下位置。在子區(qū)間中，我們既可以繼續(xù)使用二分查找，也可以使用線性的查找方式。（這里的結(jié)局方案顯然是使用了二分的方式）