题目一：轮转数组
三种思路，时间复杂度越优越好
第一种思路: 直接暴力求解，空间复杂度为o(1),但时间复杂度为o(n^2)
#include <stdio.h>
void rotate(int* nums, int k, int len);
int main()
{int arr[] = { 1,2,3,4,5,6,7 };rotate(arr, 3, sizeof(arr) / sizeof(int));for (int i = 0; i < sizeof(arr) / sizeof(int); i++){printf("%d ", arr[i]);}}
void rotate(int* nums, int k, int len)
{if (k > len){k %= len;}int i = 0, j = 0;for (i = 0; i < k; i++){int tmp = nums[len - 1];for (j = len - 1; j > 0; j--){nums[j] = nums[j - 1];}nums[j] = tmp;}
}
第二种思路: 调用库函数，时间复杂度o(n),但空间复杂度为o(n)
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
void rotate(int* nums, int k, int len);
int main()
{int arr[] = { 1,2,3,4,5,6,7 };rotate(arr, 3, sizeof(arr) / sizeof(int));for (int i = 0; i < sizeof(arr) / sizeof(int); i++){printf("%d ", arr[i]);}}
void rotate(int* nums, int k, int len)
{if (k > len){k %= len;}int* tmp = (int*)malloc(sizeof(int) * len);if (tmp == NULL){perror("malloc failed\n");return 1;}memcpy(tmp, nums + len - k, sizeof(int) * k);memcpy(tmp + k, nums, sizeof(int) * (len - k));memcpy(nums, tmp, sizeof(int) * len);free(tmp);
}
第三种思路: 翻转三次，时间复杂度0(n),空间复杂度o(1)，最优解
#include <stdio.h>
void rotate(int* nums, int k, int len);
void Reverse(int* arr, int left, int right);
int main()
{int arr[] = { 1,2,3,4,5,6,7 };rotate(arr, 3, sizeof(arr) / sizeof(int));for (int i = 0; i < sizeof(arr) / sizeof(int); i++){printf("%d ", arr[i]);}}
void Reverse(int* arr, int left, int right)
{while (left < right){int tmp = arr[left];arr[left] = arr[right];arr[right] = tmp;left++;right--;}
}
void rotate(int* nums, int k, int len)
{if (k > len){k %= len;}Reverse(nums, 0, len - k - 1);Reverse(nums, len - k, len - 1);Reverse(nums, 0, len - 1);
}
题目二：消失的数字
两种思路，时间复杂度要求o(n)
第一种思路: 直接公式，求和（等差数列）
#include <stdio.h>
int missingNumber(int* nums, int numsSize);
int main()
{int arr[] = { 0,2,3 };int ret = missingNumber(arr, 3);printf("%d", ret);return 0;
}
int missingNumber(int* nums, int numsSize)
{int sum = (0 + numsSize) * (numsSize + 1) / 2;for (int i = 0; i < numsSize; i++){sum -= nums[i];}return sum;
}
第二种思路: 异或（可参考往期博客单身狗）
#include <stdio.h>
int missingNumber(int* nums, int numsSize);
int main()
{int arr[] = { 0,2,3 };int ret = missingNumber(arr, 3);printf("%d", ret);return 0;
}
int missingNumber(int* nums, int numsSize)
{int val = 0;for (int i = 0; i < numsSize; i++){val ^= nums[i];}for (int i = 0; i <= numsSize; i++){val ^= i;}return val;
}
)
