[LeetCode][JavaScript]Search for a Range

Search for a Range 

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].

https://leetcode.com/problems/search-for-a-range/


给定有序的数列,要求找出目标,要求时间复杂度O(logn),二分法。

 1 /**
 2  * @param {number[]} nums
 3  * @param {number} target
 4  * @return {number[]}
 5  */
 6 var searchRange = function(nums, target) {
 7     var len = nums.length, start = 0; end = len - 1;
 8     while(start <= end){
 9         var index = parseInt((start + end) / 2);
10         if(nums[index] === target){
11             start = end = index;
12             while(start - 1 >= 0 && nums[start - 1] === target) start--;
13             while(end + 1 <= len - 1 && nums[end + 1] === target) end++;
14             return [start, end];
15         }else if(nums[index] > target){
16             end = index - 1;
17         }else if(nums[index] <target){
18             start = index + 1;
19         }
20     }
21     return [-1, -1];
22 };
 
原文地址:https://www.cnblogs.com/Liok3187/p/4803491.html