每日算法——leetcode系列

Search for a Range

Difficulty: Medium

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].

class Solution {public:    vector
    
      searchRange(vector
     
      & nums, int target) {            }};
     
    

翻译

搜索(目标的)所在范围

难度系数：中等

给定一个有序整数数组，找出给定值在其中的起始与结束索引。

算法的时间复杂度必须为O(logn)。

如果数组中没有指定值，返回[-1, -1]。

例如，给定[5, 7, 7, 8, 8, 10]，目标值为8，返回[3, 4]。

思路

对于有序数组， 查找可以用二分查找

代码

class Solution {public:    vector
    
      searchRange(vector
     
      & nums, int target) {        int n = (int)nums.size();        int pos = binarySearch(nums, 0, n-1, target);        vector
      
        result;        int low = -1, high = -1;        if (pos >= 0){            low = pos;            int l = low;            while (l >= 0) {                low = l;                l = binarySearch(nums, 0, low - 1, target);            }                        high = pos;            int h = high;            while (h >= 0){                high = h;                h = binarySearch(nums, high + 1, n-1, target);            }        }                result.push_back(low);        result.push_back(high);        return result;    }    private:    int binarySearch(vector
       
         nums, int low, int high, int target){                while (low <= high) {            int mid = low + (high - low)/2;            if (nums[mid] == target) {                return mid;            }            if (target > nums[mid]) {                low = mid + 1;            }            if (target < nums[mid]) {                high = mid - 1;            }        }        return -1;    }};