34. Find First and Last Position of Element in Sorted Array (M)

本文介绍了一种在已排序整数数组中查找特定目标值起始和结束位置的算法，该算法的时间复杂度为O(logn)，并提供了详细的代码实现。如果目标值不在数组中，则返回[-1,-1]。

Find First and Last Position of Element in Sorted Array (M)

Given an array of integers nums sorted in ascending order, find the starting and ending position of a given targetvalue.

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].

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]

题意

在给定的递增数组中找到目标值出现的第一个位置和最后一个位置，未找到目标值则返回[-1, -1]。

思路

查找第一个位置时，通过二分法找到目标值后，先向左递归看左侧是否仍存在目标值，存在则返回递归值，不存在则返回当前值；同理可查找最后一个位置。

代码实现

class Solution {
    public int[] searchRange(int[] nums, int target) {
        // 当不存在目标值时，必然有 first = last = -1
        int first = binarySearch(nums, 0, nums.length - 1, target, true);
        int last = first == -1 ? -1 : binarySearch(nums, 0, nums.length - 1, target, false);
        return new int[]{first, last};
    }

    private int binarySearch(int[] nums, int left, int right, int target, boolean findLeft) {
        while (left <= right) {
            int mid = (left + right) / 2;
            if (target < nums[mid]) {
                right = mid - 1;
            } else if (target > nums[mid]) {
                left = mid + 1;
            } else {
                int value;
                // 每次先判断在左/右侧是否仍存在目标值
                if (findLeft) {
                    value = binarySearch(nums, left, mid - 1, target, true);
                } else {
                    value = binarySearch(nums, mid + 1, right, target, false);
                }
                return value == -1 ? mid : value;
            }
        }
        return -1;
    }
}