递归的基本思想是把打的问题转化为小的问题,因为这些问题的解决方法很多时候是一样的,所以在函数中表现出来就是函数调用它自身的情况,但递归的时间复杂度和空间复杂度比较高。在解决很多复杂的问题的时候不推荐用递归。
递归累加:#include汉诺塔:int addTo(int paraN) { int tempSum; printf("entering addTo(%d)rn", paraN); if (paraN <= 0) { printf(" return 0rn"); return 0; } else { tempSum = addTo(paraN - 1) + paraN; printf(" return %drn", tempSum); return tempSum; }// Of if }// Of addTo int clearAddTo(int paraN) { if (paraN <= 0) { return 0; } else { return clearAddTo(paraN - 1) + paraN; } } void addToTest() { int n, sum; printf("---- addToTest begins. ----rn"); n = 5; sum = addTo(n); printf("rn0 adds to %d gets %d.rn", n, sum); n = 1; sum = addTo(n); printf("rn0 adds to %d gets %d.rn", n, sum); n = -1; sum = addTo(n); printf("rn0 adds to %d gets %d.rn", n, sum); printf("---- addToTest ends. ----rn"); } int main() { addToTest(); }
#includevoid hanoi(int paraN, char paraSource, char paraDestination, char paraTransit) { if (paraN <= 0) { return; } else { hanoi(paraN - 1, paraSource, paraTransit, paraDestination); printf("%c -> %c rn", paraSource, paraDestination); hanoi(paraN - 1, paraTransit, paraDestination, paraSource); } } void hanoiTest() { printf("---- addToTest begins. ----rn"); printf("2 platesrn"); hanoi(2, 'A', 'B', 'C'); printf("3 platesrn"); hanoi(3, 'A', 'B', 'C'); printf("---- addToTest ends. ----rn"); } int main() { hanoiTest(); }



