剑指offer 50 找到数组中重数的数字

最新推荐文章于 2023-06-25 21:29:22 发布

liufanlibashanxi

最新推荐文章于 2023-06-25 21:29:22 发布

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本文介绍了一种通过排序来查找数组中重复数字的算法。首先使用插入排序对数组进行排序，然后遍历数组，当发现相邻两个元素相等时，则找到了重复的数字。文章包含完整的C++代码实现。

1 先排序，再寻找

#include <iostream>
#include <vector>

using namespace std;

class Solution
{
public:
    
    //找到数组中重复的数字
    bool duplicate(vector<int> &numbers, int* duplication) {
        /*if (CheckValidity(numbers) == false) {
            return false;
        }*/
        *duplication = -1;
        SectionSort(numbers);
        int i = 0;
        bool isDup = false;
        *duplication = -1;
        for (i = 0; i < numbers.size() - 1; i++) {
            if (numbers[i] == numbers[i + 1]) {
                isDup = true;
                *duplication = numbers[i];
                
                break;
            }
        }
        return isDup;
    }
    // 检查合法性
    bool CheckValidity(vector<int> &numbers) {
        if (numbers.size() == 0) {
            return false;
        }
        for (int i = 0; i < numbers.size(); i++) {
            if (numbers[i] < 0 || numbers[i] > numbers.size() - 1) {
                return false;
            }
        }

        return true;
    }
    //插入排序
    void SectionSort(vector<int> &array) {
        if (array.size() == 0 || array.size() < 2) {
            return;
        }
        for (int i = 1; i < array.size(); i++) {
            for (int j = i-1; j >= 0; j--) {
                if (array[j] > array[j+1]) {
                    int temp = array[j];
                    array[j] = array[j+1];
                    array[j+1] = temp;
                    //swap(array[j], array[j+1]);
                }
            }
        }

    }

};

int main()
{
    int a1[] = { 5, 7 , 6, 3, 4, 9, 8};
    int a2[] = { 1, 2 , 4, 8, 4, 8, 9};
    vector<int> vec(a2, a2 + 7);
    int num2 = 0;


    Solution s;
    cout << s.duplicate(vec, &num2) << endl;
    cout << num2 << endl;
    //cout << s.GetNumberOfK2(vec, 8) << endl;

    /*for (int i = 0; i < array.size(); i++) {
        cout << array[i] << endl;
    }*/
}