问题描述

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

If target is not found in the array, return [-1, -1].

You must write an algorithm with O(log n) runtime complexity.

测试样例

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Input: nums = [5,7,7,8,8,10], target = 8
Output: [3,4]
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Input: nums = [5,7,7,8,8,10], target = 6
Output: [-1,-1]
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Input: nums = [], target = 0
Output: [-1,-1]

说明

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0 <= nums.length <= 10^5
-10^9 <= nums[i] <= 10^9
nums is a non-decreasing array.
-10^9 <= target <= 10^9

解题

思路

根据二分法实现寻找数组中某元素上下界的函数即可，注意处理边界条件。

补充：

  1. 时间复杂度 O(logn)

代码

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class Solution {
    public int[] searchRange(int[] nums, int target) {
        int beg = lower_bound(nums, target);
        int end = upper_bound(nums, target);
        
        return beg == end ? new int[]{-1, -1} : new int[]{beg, end - 1};
    }
    
    /**
     * 寻找下界，[beg, end)
     */
    public int lower_bound(int[] nums, int target) {
        int l = 0, r = nums.length;
        
        while(l < r) {
            int mid = (l + r) / 2;
            
            if(nums[mid] < target) {
                l = mid + 1;
            }
            else {
                r = mid;
            }
        }
        
        return l;
    }
    
    /**
     * 寻找上界，[beg, end)
     */
    public int upper_bound(int[] nums, int target) {
        int l = 0, r = nums.length;
        
        while(l < r) {
            int mid = (l + r) / 2;
            
            if(nums[mid] <= target) {
                l = mid + 1;
            }
            else {
                r = mid;
            }
        }
        
        return l;
    }
}