如题：
利用数组的左右指针，从两端往两头遍历数组，若左指针遇到偶数，右指针遇到奇数则奇偶替换，晓得左右指针不能挪动地位
#include<stdio.h>#include<string.h>void swap(int* left, int* right)//办法一{ while (left < right) { if ((*left) % 2 == 1) { left++; } if ((*right) % 2 == 0) { right--; } else { int temp = 0; temp = *left; *left = *right; *right = temp; } }}void swap1(int* left, int* right)//办法二{ while (left < right) { while ((*left) % 2 == 1) { left++; } while ((*right) % 2 == 0) { right--; } while (left < right) { int temp = 0; temp = *left; *left = *right; *right = temp; } }}void print(int* arr,int len){ int i = 0; for ( i = 0; i < len; i++) { printf("%3d", arr[i]); }}int main(){ int arr[] = { 1,6,5,2,7,8,10,4,9 }; int len = sizeof(arr) / sizeof(arr[0]); //swap(arr, arr + len - 1); swap1(arr, arr + len - 1); print(arr, len);}正文：当数组全为奇数或偶数时，应加上约束条件left<right避免指针越界拜访数组元素