Sword refers to the offer series source code-the number of occurrences in the array is more than half

Topic 1370: Digital time limit for more than half occurrences in an array: 1 seconds Memory limit: 32 Mega Special: No submission: 2844 Resolution: 846 title Description: There is a number in the array that appears more than half the length of the array, please find this number. For example, enter an array of length 9 {1,2,3,2,2,2,5,4,2}. Since the number 2 appears in the array 5 times, which exceeds half the length of the array, the output is 2. Input: Each test case consists of 2 rows: The first line enters an integer n (1<=n<=100000) that represents the number of elements in the array. The second line enters n integers, representing each element in the array, and the range of n integers is [1,1000000000]. Output: corresponding to each test case, the output occurs more than half the length of the array, if there is no output-1. Sample input: 91 2 3 2 2 2 5 4 2 sample output: 2

The most intuitive solution:

Sort the array, select the median, and then compare whether the value is equal to the median and the total is greater than N/2.

#include <iostream> #include <stdio.h>using namespace std;//find base position int adjust (int a[],int left,int right) {int    x = A[left];        while (Left<right) {while (left<right&&a[right]>x) {right--;            } if (left<right) {A[left] = A[right];        left++;        } while (left<right&&a[left]<x) {left++;            } if (left<right) {a[right] = A[left];        right--;    }} A[left] = x; return left;}         Recursion (so that the base position interval is one number) void Quick_sort (int a[],int left,int right) {if (left<right) {int i = adjust (a,left,right);        Quick_sort (a,left,i-1);    Quick_sort (A,i+1,right);    }}int Middle (int *a,int length) {int x = A[LENGTH/2];    int count = 0;        for (int i=0;i<length;i++) {if (X==a[i]) {count++;    }} if (Count*2>length) {return x; } return-1;}    int main () {int n;   while (scanf ("%d", &n)!=eof) {     int* a = new int[n];    for (int i=0;i<n;i++) {scanf ("%d", &a[i]);    } quick_sort (a,0,n-1);    printf ("%d\n", Middle (a,n)); } return 0;}

Sword refers to the offer series source code-the number of occurrences in the array is more than half