Find maximum repeating number

给你一个长度是n的整数数组，数组中数字的范围是 [0, k-1] ,  k 满足 k<=n。找到该数组中，重复出现次数最多的数字（当出现次数相同时，找出值最大的那个）。例如 k=10, arr[] = {1, 2, 2, 2, 0, 2, 0, 2, 3, 8, 0, 9, 2, 3}，那么符合条件的是数字 2 。请给出一个时间复杂度是O(n)，空间复杂度是O(1)的算法。

思路：

假如给定arr[] = {2, 3, 3, 5, 3, 4, 1, 7}, k = 8, n = 8。那么，

1. Iterate though input array arr[], for every element arr[i], increment arr[arr[i]%k] by k (arr[] becomes {2, 11, 11, 29, 11, 12, 1, 15 })
2. Find the maximum value in the modified array (maximum value is 29). Index of the maximum value is the maximum repeating element (index of 29 is 3).
3. If we want to get the original array back, we can iterate through the array one more time and do arr[i] = arr[i] % k where i varies from 0 to n-1.

具体代码如下：

#include<iostream>using namespace std; // Returns maximum repeating element in arr[0..n-1].// The array elements are in range from 0 to k-1int maxRepeating(int* arr, int n, int k){    // Iterate though input array, for every element    // arr[i], increment arr[arr[i]%k] by k    for (int i = 0; i< n; i++)        arr[arr[i]%k] += k;     // Find index of the maximum repeating element    int max = arr[0], result = 0;    for (int i = 1; i < n; i++)    {        if (arr[i] > max)        {            max = arr[i];            result = i;        }    }     /* Uncomment this code to get the original array back       for (int i = 0; i< n; i++)          arr[i] = arr[i]%k; */     // Return index of the maximum element    return result;} // Driver program to test above functionint main(){    int arr[] = {2, 3, 3, 5, 3, 4, 1, 7};    int n = sizeof(arr)/sizeof(arr[0]);    int k = 8;     cout << "The maximum repeating number is " <<         maxRepeating(arr, n, k) << endl;     return 0;}