From 8b790180a1ec2f6508e7367f43390a11516ec0d6 Mon Sep 17 00:00:00 2001
From: pmz9tlfpr <23184979050@qq.com>
Date: Sun, 25 Jun 2023 23:33:39 +0800
Subject: [PATCH] =?UTF-8?q?Add=20=E4=BC=97=E6=95=B0=E9=87=8D=E6=95=B0?=
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---
 众数重数 | 44 ++++++++++++++++++++++++++++++++++++++++++++
 1 file changed, 44 insertions(+)
 create mode 100644 众数重数

diff --git a/众数重数 b/众数重数
new file mode 100644
index 0000000..6c194a7
--- /dev/null
+++ b/众数重数
@@ -0,0 +1,44 @@
+#include <stdio.h>
+//求解结果表示
+int num;							//全局变量,存放众数
+int maxcnt = 0;						//全局变量,存放重数
+void split(int a[], int low, int high, int& mid, int& left, int& right)
+//以a[low..high]中间的元素为界限,确定为等于a[mid]元素的左、右位置left和right
+{
+	mid = (low + high) / 2;
+	for (left = low; left <= high; left++)
+		if (a[left] == a[mid])
+			break;
+	for (right = left + 1; right <= high; right++)
+		if (a[right] != a[mid])
+			break;
+	right--;
+}
+void Getmaxcnt(int a[], int low, int high)
+{
+	if (low <= high)					//a[low..high]序列至少有1个元素
+	{
+		int mid, left, right;
+		split(a, low, high, mid, left, right);
+		int cnt = right - left + 1;			//求出a[mid]元素的重数
+		if (cnt > maxcnt)				//找到更大的重数
+		{
+			num = a[mid];
+			maxcnt = cnt;
+		}
+		Getmaxcnt(a, low, left - 1);		//左序列递归处理
+		Getmaxcnt(a, right + 1, high);	//右序列递归处理
+	}
+}
+int main()
+{
+	int a[] = { 1,2,2,2,3,3,5,6,6,6,6 };
+	int n = sizeof(a) / sizeof(a[0]);
+	printf("求解结果\n");
+	printf("  递增序列: ");
+	for (int i = 0; i < n; i++)
+		printf("%d ", a[i]);
+	printf("\n");
+	Getmaxcnt(a, 0, n - 1);
+	printf("  众数: %d, 重数: %d\n", num, maxcnt);
+}