为什么我不能从声明为返回字符串的函数返回整数？

Question

class Solution {
public:
    string kthLargestNumber(vector<string>& nums, int k) {
        priority_queue<int>q;
        for(int i=0;i<nums.size();i++)
        {
            q.push(stoi(nums[i]));
        }
        while(!q.empty())
        {
                k--;
                if(k==0)
                {
                    return q.top();
                }
                q.pop();
        }
        return 0;   
    }
};

我试图解决 Leetcode 1985. 找出数组中第 K 个最大整数的问题。但我在第 14 行遇到编译错误 (

return q.top();

) 我不知道为什么会发生这种情况。

问题链接

Answer 1

只要看一下代码，答案就很简单了。该函数应返回

std::string

，而您则返回 和

int

。您现在可能想通过使用

to_string

函数并将

int

转换为

std:string

来解决问题。

但是，这是行不通的。

该解决方案应该而且必须适用于字符串。

Constraints:
    1 <= k <= nums.length <= 104
    1 <= nums[i].length <= 100
    nums[i] consists of only digits.
    nums[i] will not have any leading zeros.

很明显，你不能在

int

中存储100位长的数字。因此，您还必须使用字符串排序。

但这会让事情变得困难，因为你需要采用所谓的自然排序。

示例：查看 2,3 和 200，然后，如果您比较字符串，3 将大于 200。这就是他们想要从您那里得到的。自然排序。

有很多可能的解决方案。这是其中之一：

#include <iostream>
#include <string>
#include <cctype>
#include <algorithm>
#include <vector>
using namespace std::string_literals;

// FUNCTOR for Natural sort function. Limited to signed char strings
struct NaturalSort {

    bool operator ()(const std::string& s1, const std::string& s2) {

        // We want to iterate over both strings
        const char* i1 = s1.data(); // Get a pointer to the beginning of string 1
        const char* i2 = s2.data(); // Get a pointer to the beginning of string 2

        // Result of comparison. 0 means equal, positive means s1>s2, negative means s1<s2
        int result = 0;
        // As long as both characters/numbers are equal and there are still characters in the string
        do {
            // If we found a digit at the same position in both strings
            if (std::isdigit(*i1) and std::isdigit(*i2)) {

                std::size_t pos1{}, pos2{}; // This will later indicate how many digits we read
                // Convert the strings to numbers and compare them
                result = std::stoi(i1, &pos1) - std::stoi(i2, &pos2);
                // Set pointers to the position after the number
                i1 += pos1; i2 += pos2;
            }
            else {
                // Normal character
                result = (*i1 - *i2);
                ++i1; ++i2;
            }
        } while ((result == 0) and (*i1 != 0 or *i2 != 0));
        // Let's limit the output to -1, 0 ,1
        return result < 0;
    }
};


class Solution{
public:
    std::string kthLargestNumber(std::vector<std::string>& nums, int k) {
        std::sort(nums.begin(), nums.end(), NaturalSort());
        return nums[nums.size()-k];
    }
};

// Driver / test code
int main() {
    std::vector<std::string> test{ "2"s,"3"s,"20"s,"30"s,"200"s,"300"s };
    Solution solution{};
    for (int k = 1; k <= test.size(); ++k) 
        std::cout << "With k=" << k << " the result would be: " << solution.kthLargestNumber(test, k) << '\n';
}