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d26e811e10
* fix the bugs of C code. * Add a header figure. * Improve the definition of tree node height.
77 lines
1.6 KiB
C
77 lines
1.6 KiB
C
/**
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* File: recursion.c
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* Created Time: 2023-09-09
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* Author: Gonglja (glj0@outlook.com)
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*/
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#include "../utils/common.h"
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/* 递归 */
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int recur(int n) {
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// 终止条件
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if (n == 1)
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return 1;
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// 递:递归调用
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int res = recur(n - 1);
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// 归:返回结果
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return n + res;
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}
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/* 使用迭代模拟递归 */
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int forLoopRecur(int n) {
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int stack[1000]; // 借助一个大数组来模拟栈
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int top = -1; // 栈顶索引
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int res = 0;
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// 递:递归调用
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for (int i = n; i > 0; i--) {
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// 通过“入栈操作”模拟“递”
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stack[1 + top++] = i;
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}
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// 归:返回结果
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while (top >= 0) {
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// 通过“出栈操作”模拟“归”
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res += stack[top--];
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}
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// res = 1+2+3+...+n
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return res;
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}
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/* 尾递归 */
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int tailRecur(int n, int res) {
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// 终止条件
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if (n == 0)
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return res;
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// 尾递归调用
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return tailRecur(n - 1, res + n);
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}
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/* 斐波那契数列:递归 */
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int fib(int n) {
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// 终止条件 f(1) = 0, f(2) = 1
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if (n == 1 || n == 2)
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return n - 1;
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// 递归调用 f(n) = f(n-1) + f(n-2)
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int res = fib(n - 1) + fib(n - 2);
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// 返回结果 f(n)
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return res;
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}
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/* Driver Code */
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int main() {
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int n = 5;
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int res;
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res = recur(n);
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printf("\n递归函数的求和结果 res = %d\n", res);
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res = forLoopRecur(n);
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printf("\n使用迭代模拟递归求和结果 res = %d\n", res);
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res = tailRecur(n, 0);
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printf("\n尾递归函数的求和结果 res = %d\n", res);
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res = fib(n);
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printf("\n斐波那契数列的第 %d 项为 %d\n", n, res);
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return 0;
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}
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