[leetcode] 34. Find the first and last positions of elements in the sorted array

34、 Sort the first and last positions of the elements in the array

subject ：

Give you an array of integers in non decreasing order nums, And a target value target. Please find out the start position 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].

You must design and implement a time complexity of O(log n) The algorithm to 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

Their thinking ：

The intuitive idea must be to traverse from front to back . Record the first and last encounter with two variables target The subscript , But the time complexity of this method is O(n), Didn't make use of Array ascending Conditions of arrangement .

Two points search

Because the array has been sorted , So the whole array is monotonically increasing , We can use Dichotomy To speed up the search process .

consider target Start and end positions , In fact, what we're looking for is in the array 「 The first is equal to target The location of 」（ Write it down as leftIdx） and 「 The first one is greater than target Minus one 」（ Write it down as rightIdx）.

Binary search , seek leftIdx That is to find the first in the array Greater than or equal to target The subscript , seek rightIdx That is to find the first in the array Greater than target The subscript , Then subtract the subscript by one .

The judgment conditions of the two are different , For code reuse , We define binarySearch(nums, target, lower) It means that nums Binary search in array target The location of , If lower by true, Then find the first Greater than or equal to target The subscript , Otherwise, find the first greater than target The subscript .

Last , because target May not exist in the array , So we need to recheck the two subscripts we get leftIdx and rightIdx, See if they meet the requirements , If the conditions are met, return [leftIdx,rightIdx], Go back if you don't agree [-1,-1][−1,−1].

Reference ideas ：

class Solution {
    
    public int[] searchRange(int[] nums, int target) {
    
        int leftIdx = binarySearch(nums, target, true);
        int rightIdx = binarySearch(nums, target, false) - 1;
        if (leftIdx <= rightIdx && rightIdx < nums.length && nums[leftIdx] == target && nums[rightIdx] == target) {
    
            return new int[]{
    leftIdx, rightIdx};
        } 
        return new int[]{
    -1, -1};
    }

    public int binarySearch(int[] nums, int target, boolean lower) {
    
        int left = 0, right = nums.length - 1, ans = nums.length;
        while (left <= right) {
    
            int mid = (left + right) / 2;
            if (nums[mid] > target || (lower && nums[mid] >= target)) {
    
                right = mid - 1;
                ans = mid;
            } else {
    
                left = mid + 1;
            }
        }
        return ans;
    }
}