标签： C++ 算法 LeetCode 数组 二分查找

每日算法——leetcode系列

问题 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<int> searchRange(vector<int>& 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
由于有重复的值，如果二分法找到目标，则分两部分继续二分查找
如果没找到，返回[-1, -1]

代码

class Solution {
public:
    vector<int> searchRange(vector<int>& nums, int target) {
        int n = (int)nums.size();
        int pos = binarySearch(nums, 0, n-1, target);

        vector<int> 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<int> 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;
    }
};