用 break 跳出循环
用 return 跳出函数
用 continue 结束本次循环开始下次循环
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- #include <> typedef struct Node { int index; struct Node *next; }JosephuNode; int Josephu(int n, int m) { int i, j; JosephuNode *head, *tail; head = tail = (JosephuNode *)malloc(sizeof(JosephuNode)); for (i = 1; i < n; ++i) { tail->index = i; tail->next = (JosephuNode *)malloc(sizeof(JosephuNode)); tail = tail->next; } tail->index = i; tail->next = head; for (i = 1; tail != head; ++i) { for (j = 1; j < m; ++j) { tail = head; head = head->next; } tail->next = head->next; printf("第%4d个出局的人是：%4d号\n", i, head->index); free(head); head = tail->next; } i = head->index; free(head); return i; } int main { int n, m; scanf("%d%d", &n, &m); printf("最后胜利的是%d号！\n", Josephu(n, m)); system("pause"); return 0; }">设编号为1，2，… n的n个人围坐一圈，约定编号为k（1数组实现： #include <> #include <> int Josephu(int n, int m) { int flag, i, j = 0; int *arr = (int *)malloc(n * sizeof(int)); for (i = 0; i < n; ++i) arr[i] = 1; for (i = 1; i < n; ++i) { flag = 0; while (flag < m) { if (j == n) j = 0; if (arr[j]) ++flag; ++j; } arr[j - 1] = 0; printf("第%4d个出局的人是：%4d号\n", i, j); } free(arr); return j; } int main { int n, m; scanf("%d%d", &n, &m); printf("最后胜利的是%d号！\n", Josephu(n, m)); system("pause"); return 0; } 链表实现： #include <> #include <> typedef struct Node { int index; struct Node *next; }JosephuNode; int Josephu(int n, int m) { int i, j; JosephuNode *head, *tail; head = tail = (JosephuNode *)malloc(sizeof(JosephuNode)); for (i = 1; i < n; ++i) { tail->index = i; tail->next = (JosephuNode *)malloc(sizeof(JosephuNode)); tail = tail->next; } tail->index = i; tail->next = head; for (i = 1; tail != head; ++i) { for (j = 1; j < m; ++j) { tail = head; head = head->next; } tail->next = head->next; printf("第%4d个出局的人是：%4d号\n", i, head->index); free(head); head = tail->next; } i = head->index; free(head); return i; } int main { int n, m; scanf("%d%d", &n, &m); printf("最后胜利的是%d号！\n", Josephu(n, m)); system("pause"); return 0; }
- java实现斐波那契数列，现在要求输入一个整数n，请你输出斐波那契数列的第n项。 n public class Solution_feibonaqi { public int Fibonacci(int n) { int result[] = { 0, 1 }; if (n < 2) { return result[n]; } int f0 = 0; int f1 = 1; int f2 = 0; for (int i = 2; i <= n; i++) { f2 = f1 + f0; f0 = f1; f1 = f2; } return f2; } public static void main(String[] args) { Scanner sc = new Scanner; int n = ; Solution_feibonaqi fei = new Solution_feibonaqi; ((n)); } }
- return & finally 执行顺序 这是我读到的最合理的解释
- while中的break和continue