Java implementation looks for elements in the sorted array in the first and last position LeetCode 34

34. Find a sorted array elements in the first and last position

Nums Given an array of integers arranged in ascending order, according to the target and one target value. Identify a given target start and end positions in the array.

Your time complexity of the algorithm must be O (log n) level.

If the target is not present 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]

Source: stay button (LeetCode)
link: https: //leetcode-cn.com/problems/find-first-and-last-position-of-element-in-sorted-array
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PS: Let me talk about a simple (using the powerful Java JDK)

class Solution {
    public int[] searchRange(int[] nums, int target) {
        int i = Arrays.binarySearch(nums, target);
        if (i<0) return new int[]{-1,-1};
        int begin = i;
        int end = i;
        while (begin-1>=0 && nums[begin-1]==target) {begin--;}
        while (end+1<nums.length && nums[end+1]==target) {end++;}
        return new int[]{begin,end};
    }
}

PS: dichotomy

class Solution {
    public int[] searchRange(int[] nums, int target) {
        int i = 0, j = nums.length;
        int mid = (i + j) / 2;
        int p = -1;
        while (i < j) {  //找到target的位置
            if (nums[mid] == target) {
                p = mid;
                break;
            }
            if (nums[mid] > target) {
                if (j == mid) break;
                j = mid;
                mid = (i + j) / 2;
            } else {
                if (i == mid) break;
                i = mid;
                mid = (i + j) / 2;
            }
        }

        if (p == -1) {
            return new int[]{-1, -1};
        } else {  //在target所在位置向前向后查找
            int a = p, b = p;
            while (a > 0 && nums[a - 1] == target) a--;
            while (b < nums.length - 1 && nums[b + 1] == target) b++;
            return new int[]{a, b};
        }
    }
}