# 兩次最大出現之間的最小距離 > 原文： [https://www.geeksforgeeks.org/minimum-distance-between-two-occurrences-of-maximum/](https://www.geeksforgeeks.org/minimum-distance-between-two-occurrences-of-maximum/) 您將獲得一個 n 元素數組，其基本條件是最大元素的出現不止一次。 您必須找到最大值之間的最小距離。 （n > = 2）。 例子： ``` Input : arr[] = {3, 5, 2, 3, 5, 3, 5} Output : Minimum Distance = 2 Explanation : Greatest element is 5 and its index are 1, 4 and 6\. Resulting minimum distance of 2 from position 4 to 6. Input : arr[] = {1, 1, 1, 1, 1, 1} Output : Minimum Distance = 1 Explanation : Greatest element is 1 and its index are 0, 1, 2, 3, 4 and 5\. Resulting minimum distance of 1. ``` **基本方法**在 `O(n^2)`中運行。 首先，我們找到最大元素。 然后，對于等于最大元素的每個元素，我們找到最接近的最大元素。 一個有效的**解決方案**在數組的單個遍歷中完成了我們的工作。 我們初始化 maximum_element = arr [0]，min_distance = n 和 index =0。之后，對于每個元素，我們應檢查元素是否等于，大于或小于最大元素。 根據三種情況，我們有以下選擇： * **情況 a**：如果 element 等于 maximum_element，則更新 min_dis = min（min_dis，（i-index））并更新 index = i; * **情況 b**：如果元素大于 maximum_element，則更新 maximum_element = arr [i]，索引= i，min_dis = n。 * **情況 c**：如果 element 小于 maximum_element，則迭代到下一個元素。 ## C ``` // C program to find Min distance? // of maximum element #include<bits/stdc++.h> using namespace std; //function to return min distance int minDistance (int arr[], int n) { ????int maximum_element = arr[0]; ????int min_dis = n; ????int index = 0; ????for (int i=1; i<n; i++) ????{ ????????// case a ????????if (maximum_element == arr[i]) ????????{ ????????????min_dis = min(min_dis, (i-index)); ????????????index = i; ????????} ????????// case b ????????else if (maximum_element < arr[i]) ????????{ ????????????maximum_element = arr[i]; ????????????min_dis = n; ????????????index = i; ????????} ????????// case c ????????else ????????????continue; ????} ????return min_dis; } // driver program int main() { ????int arr[] = {6, 3, 1, 3, 6, 4, 6}; ????int n = sizeof(arr)/sizeof(arr[0]); ????cout << "Minimum distance = "? ?????????<< minDistance(arr, n); ????return 0; }? ``` ## Java ```java // Java program to find Min distance? // of maximum element class GFG? { ????// function to return min distance ????static int minDistance (int arr[], int n) ????{ ????????int maximum_element = arr[0]; ????????int min_dis = n; ????????int index = 0; ????????for (int i=1; i<n; i++) ????????{ ????????????// case a ????????????if (maximum_element == arr[i]) ????????????{ ????????????????min_dis = Math.min(min_dis, (i-index)); ????????????????index = i; ????????????} ????????????// case b ????????????else if (maximum_element < arr[i]) ????????????{ ????????????????maximum_element = arr[i]; ????????????????min_dis = n; ????????????????index = i; ????????????} ????????????// case c ????????????else ????????????????continue; ????????} ????????return min_dis; ????} ????// Driver code ????public static void main (String[] args) ????{ ????????int arr[] = {6, 3, 1, 3, 6, 4, 6}; ????????int n = arr.length; ????????System.out.print("Minimum distance = "+minDistance(arr, n)); ????} } // This code is contributed by Anant Agarwal. ``` ## Python3 ```py # Python3 program to find Min? # distance of maximum element # Function to return min distance def minDistance (arr, n): ????maximum_element = arr[0] ????min_dis = n ????index = 0 ????for i in range(1, n): ????????# case a ????????if (maximum_element == arr[i]): ????????????min_dis = min(min_dis, (i - index)) ????????????index = i ????????# case b ????????elif (maximum_element < arr[i]): ????????????maximum_element = arr[i] ????????????min_dis = n ????????????index = i ????????# case c ????????else: ????????????continue ????return min_dis # driver program arr = [6, 3, 1, 3, 6, 4, 6] n = len(arr) print("Minimum distance =", minDistance(arr, n)) # This code is contributed by Smitha Dinesh Semwal ``` ## C# ```cs // C# program to find Min distance? // of maximum element using System; class GFG { ????// function to return min distance ????static int minDistance (int []arr, int n) ????{ ????????int maximum_element = arr[0]; ????????int min_dis = n; ????????int index = 0; ????????for (int i = 1; i < n; i++) ????????{ ????????????// case a ????????????if (maximum_element == arr[i]) ????????????{ ????????????????min_dis = Math.Min(min_dis,? ????????????????????????????????(i - index)); ????????????????index = i; ????????????} ????????????// case b ????????????else if (maximum_element < arr[i]) ????????????{ ????????????????maximum_element = arr[i]; ????????????????min_dis = n; ????????????????index = i; ????????????} ????????????// case c ????????????else ????????????????continue; ????????} ????????return min_dis; ????} ????// Driver code ????public static void Main () ????{ ????????int []arr = {6, 3, 1, 3, 6, 4, 6}; ????????int n = arr.Length; ????????Console.WriteLine("Minimum distance = " ????????????????????????+ minDistance(arr, n)); ????} } // This code is contributed by vt_m. ``` ## PHP ```php <?php // PHP program to find Min distance? // of maximum element //function to return min distance function minDistance($arr, $n) { ????$maximum_element = $arr[0]; ????$min_dis = $n; ????$index = 0; ????for ($i = 1; $i < $n; $i++) ????{ ????????// case a ????????if ($maximum_element == $arr[$i]) ????????{ ????????????$min_dis = min($min_dis, ($i - $index)); ????????????$index = $i; ????????} ????????// case b ????????else if ($maximum_element < $arr[$i]) ????????{ ????????????$maximum_element = $arr[$i]; ????????????$min_dis = $n; ????????????$index = $i; ????????} ????????// case c ????????else ????????????continue; ????} return $min_dis; } ????// Driver Code ????$arr = array(6, 3, 1, 3, 6, 4, 6); ????$n = count($arr); ????echo "Minimum distance = " ??????????.minDistance($arr,$n);? // This code is contributed by Sam007 ?> ``` Output : ``` Minimum distance = 2 ``` * * * * * *