[binary search medium] leetcode 34. find the first and last positions of elements in the sorted array

LeetCode 34. Find the first and last positions of the elements in the sort array

https://leetcode.cn/problems/find-first-and-last-position-of-element-in-sorted-array/comments/
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]

Solution

Two points + Traverse left and right (Java), Time complexity ：$O(logn+n)$：

class Solution {
    
    public int[] searchRange(int[] a, int target) {
    
        int[] ans = {
    -1, -1};
        int l = 0, r = a.length - 1, mid = 0;
         //  Find it by two points first  target  The corresponding number 
        while (l <= r) {
    
            mid = (l + r) / 2;
            if (a[mid] == target) {
    
                ans[0] = mid;
                ans[1] = mid;
                break;
            } else if (a[mid] < target) {
    
                l = mid + 1;
            } else {
    
                r = mid - 1;
            }
        }
		//  towards  target  The left side and the right side are traversed respectively to find equal to  target  Minimum index and maximum index 
        for (int i = mid - 1; i >= 0; i--) {
    
            if (a[i] != target) {
    
                break;
            }
            ans[0] = i;
        }
        for (int i = mid + 1; i < a.length; i++) {
    
            if (a[i] != target) {
    
                break;
            }
            ans[1] = i;
        }
        return ans;
    }
}

Find the maximum and minimum subscript in two points (C) , Time complexity ：$O(logn)$ ：
searchInsert() Function directly Copy It was written before LeetCode 35. Search the code of the insertion location , The code will find equal to target The minimum subscript of the value .

int searchInsert(int* a, int n, int target){
    
    int l = 0, r = n - 1;
    int ans = n;
    int mid;
    while (l <= r) {
    
        mid = (l + r) / 2;
        if (a[mid] >= target) {
    
            ans = mid;
            r = mid - 1;
        } else {
    
            l = mid + 1;
        }
    }
    return ans;
}

int* searchRange(int* nums, int numsSize, int target, int* returnSize){
    
    int *reslut = (int*) malloc(sizeof(int) * 2);
    *returnSize = 2; //  This must be added , No error report 
    reslut[0] = reslut[1] = -1;
    //  Find insert  target  Minimum index of location 
    int left_idx = searchInsert(nums, numsSize, target);
    //  If the minimum subscript is less than the table length and the number corresponding to the minimum index found  == target 
    if (left_idx < numsSize && nums[left_idx] == target) {
    
        reslut[0] = left_idx;
        //  Find insert  (target + 1)  Minimum index value  - 1
        // (target + 1)  Minimum index value of  - 1  Is equal to  target  Maximum index value for 
        reslut[1] = searchInsert(nums, numsSize, target + 1) - 1;
    }
    return reslut;
}