LeetCode：34. Find the first and last positions of elements in a sorted array

Given an array of integers in ascending order nums, And a target value target. Find the start and end position of the given target value in the array .

If the target value does not exist in the array target, return [-1, -1].

Advanced ：

• You can design and implement time complexity of O(log n) Does the algorithm solve this problem ？

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]

Example 3：

 Input ：nums = [], target = 0
 Output ：[-1,-1]

Tips ：

• 0 <= nums.length <= 10^5
• -10^9 <= nums[i] <= 10^9
• nums It is a group of non decreasing numbers
• -10^9 <= target <= 10^9

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Their thinking

1. First, judge that the array is not empty and this element exists , And then use indexOf and lastIndexOf Method to find the first and last positions of the element

Code

/** * @param {number[]} nums * @param {number} target * @return {number[]} */
var searchRange = function(nums, target) {
    
    if(nums !== [] && nums.indexOf(target) !== -1) {
    
        let left = nums.indexOf(target)
        let right = nums.lastIndexOf(target)
        return [left, right]
    }else {
    
        return [-1, -1]
    }
};