【LeetCode】Search for a Range

问题描述

Given a sorted array of integers, find the starting and ending position of a given target value.
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].
For example,
Given [5, 7, 7, 8, 8, 10] and target value 8,
return [3, 4].
意：找到排序数组中某个数出现的范围

算法思想

二分查找到目标数后向前后顺序查找

算法实现

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

算法时间

算法时间为二分查找时间
T(n)=O(logn)
但是如果所有数都相同时，查找时间为O(1)，顺序查找时间为O(n)，所以最坏情况为O(n)

演示结果

public static void main(String[] args) {
        int result[] = new int[] { -1, -1 };
        int[] arr = { 1, 23, 34, 34, 34, 34, 34, 54, 65, 112 };

        result = searchRange(arr, 65);
        System.out.println(result[0] + ":" + result[1]);

    }

8:8