递归算法能不能转化为非递归算法：算法单递归减而治之
递归算法能不能转化为非递归算法：算法单递归减而治之#include <stdio.h> int Sum(int arr[] int size) { if(size<1) return 0; return arr[size-1] Sum(arr size-1); // 单递归，线性递归 } int main() { int arr[] = {1 2 3 4 5}; int n = sizeof arr / sizeof *arr; printf("%d\n" Sum(arr n-1)); getchar(); }线性递归：demo:The control of a large force is the same principle as the control of a few men: it is merely a ques
迭代乃人工，递归方神勇。
To iterate is human to recurse divine.
（神通的原因在于编译器在背后利用函数栈帧的数据结构。）
凡治众如治寡，分数是也。
The control of a large force is the same principle as the control of a few men: it is merely a question of dividing up their numbers.
分数，可以减而治之（Decrease and conquer），也可以分而治之（Divide and conquer）。
1 减而治之
demo:
#include <stdio.h>
int Sum(int arr[] int size)
{
if(size<1)
return 0;
return arr[size-1] Sum(arr size-1); // 单递归，线性递归
}
int main()
{
int arr[] = {1 2 3 4 5};
int n = sizeof arr / sizeof *arr;
printf("%d\n" Sum(arr n-1));
getchar();
}
线性递归：
2 分而治之
demo:
#include <stdio.h>
int Sum(int arr[] int lo int hi)
{
if(lo==hi)
return arr[lo];
int mid = (lo hi)>>1;
return Sum(arr lo mid) Sum(arr mid 1 hi); // 双递归，二叉树形式递归
}
int main()
{
int arr[] = {1 2 3 4 5};
int n = sizeof arr / sizeof *arr;
printf("%d\n" Sum(arr 0 n-1));
getchar();
}
二叉树形式递归：
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