[LeetCode] Find First and Last Position of Element in Sorted

最新推荐文章于 2025-09-07 13:58:20 发布

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最新推荐文章于 2025-09-07 13:58:20 发布

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CC 4.0 BY-SA版权

文章标签： runtime

原文链接：https://segmentfault.com/a/1190000016327626

本文介绍了一种在已排序数组中查找指定目标值起始和结束位置的算法，该算法的时间复杂度为O(logn)，通过二分查找实现高效定位。

Problem

Given an array of integers nums sorted in ascending order, 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].

Example 1:

Input: nums = [5,7,7,8,8,10], target = 8
Output: [3,4]

Example 2:

Input: nums = [5,7,7,8,8,10], target = 6
Output: [-1,-1]

Solution

class Solution {
    public int[] searchRange(int[] nums, int target) {
        int[] res = {-1, -1};
        if (nums == null || nums.length == 0) return res;
        int start = 0, end = nums.length-1;
        while (nums[start] < nums[end]) {                   //don't be equal
            int mid = start + (end-start)/2;
            if (nums[mid] < target) {
                start = mid+1;
            } else if (nums[mid] > target) {
                end = mid-1;
            } else {                                        //once nums[mid] == target:
                if (nums[start] != nums[mid]) start++;      //move start to lower bound (first position)
                else end--;                                 //move end to higher bound (last position)
            }
        }
        
        if (nums[start] == target && nums[end] == target) {
            res[0] = start;
            res[1] = end;
        }
        return res;
    }
}