Yanghui triangle
- If we don't talk about mathematics, let's talk about its realization.
- (1) Observing the picture below, we know that the Yang Hui triangle is not a number 1 in the middle , and its value is equal to the sum of the two numbers on its shoulder . Using this rule, we can construct the code and write the middle number.
- (2) Look at the following form:
- We conclude that we can use a two-dimensional array to represent the Yang Hui triangle, and observe this graph, combined with the subscripts of the elements in the two-dimensional array, successfully found, the above law (the sum of the numbers on the shoulder) code represents!
- It is easy to figure out according to the law in the figure, for example: a[8][3] = a[7][2] + a[7][3] , and then write a loop, you can find the value of the number in each row
- (3) Regarding the law of the number 1 on both sides, it is easier to find out. They are the first and the last respectively. Then another example: In the i-th row , a[i][0] = 1, a[i][i] = 1 realizes that both sides are 1~
C language implementation
- According to the above analysis, we break down the steps:
- (1) We found that it is actually a square, and then there is only a lower triangle (right triangle as shown in the figure), so we first construct an n by n two-dimensional array, and use it to store Yang Hui's triangle.
// 打印100行杨辉三角
#include<stdio.h>
int main()
{
int yang[50][50] = {
0};
int i, j, m;
// 打印杨辉三角
for(i=0; i<50; i++)
{
// 这里只打印二维数组的杨辉三角部分没所以 j <= i
for(j=0; j<=i; j++)
{
m = yang[i][j];
printf("%d", m);
}
printf("\n");
}
return 0;
}
As shown in figure: running result
- (2) We then print out the 1 on both sides , remembering the rule of this 1 . It appears at the beginning and end of a line
// 打印100行杨辉三角
#include<stdio.h>
int main()
{
int yang[25][25] = {
0};
int i, j, m;
// 补充代码打印杨辉三角两边的 1
// 输出每行的1，位置是行首和 行尾，对应的下标是 [][0] , [][i]. i为当前行，以0开始
for(i=0; i<25; i++)
{
yang[i][0] = 1;
yang[i][i] = 1;
}
// 打印杨辉三角
for(i=0; i<25; i++)
{
for(j=0; j<=i; j++)
{
m = yang[i][j];
printf("%d", m);
}
printf("\n");
}
return 0;
}
The result:
** (3) The third step, it is a critical part of the middle of the printing, refer to the above Triangle that picture, we know the following number, equal to its shoulders on two numbers and . Then according to this rule, you can continue to write code to achieve!
After the third step is finished, all the codes are finished.
// 打印100行杨辉三角
#include<stdio.h>
int main()
{
int yang[25][25] = {
0};
int i, j, m;
// 输出每行的1，位置是行首和 行尾，对应的下标是 [][0] , [][i]. i为当前行，以0开始
for(i=0; i<25; i++)
{
yang[i][0] = 1;
yang[i][i] = 1;
// 补充代码，打印一行中间的数字
for(j=1; j<i; j++)
{
// 注意 j的取值，从每行的第二个数字，到倒数第二个数字，因为第一个最后的数字是 1固定的
// 按照数的值 是它肩上两个数的和
yang[i][j] = yang[i-1][j-1] + yang[i-1][j];
}
}
// 打印杨辉三角
for(i=0; i<25; i++)
{
for(j=0; j<=i; j++)
{
m = yang[i][j];
printf("%d ", m); // 为了格式好看，我们在这里加上两个空格，隔开数字！！！
}
printf("\n");
}
return 0;
}
Print the results to see!
- It was successful, but the result was a bit messy, and the layout in the terminal was not very good!
There are many implementation methods, but after my analysis and consideration, I wrote such a one, I hope it will be helpful to you
Python implementation
Method 1: Based on the previous C language algorithm, write it in Python syntax
- The insert method can actually be replaced by the append() method, but I want to reflect the steps in the C language algorithm, inserting the 1 on both sides, using insert, regardless of efficiency.
# 纯基本语法
yang = []
n = int(input("请输入你想打印的杨辉三角层数："))
for i in range(0, n):
temp = []
temp.insert(0, 1)
for j in range(1, i):
num = yang[i-1][j-1] + yang[i-1][j]
temp.append(num)
# 因为 insert方法会在指定位置位置增加一个 值，若原本位置上有值，那么将被后移，造成第一行有两个1， 所以加个判断
if i > 0:
temp.insert(i, 1)
yang.append(temp)
for k in yang:
print(k)
Python implementation (functional programming!)
Method 2: Use sum() and zip functions.
- This is a solution found on the Internet, it’s really amazing, I can’t think of it
def generate(numRows):
l1 = [1]
l2 = []
n = 0
while n<numRows:
l2.append(l1)
# 这里的 本质算法还是一样的 ，在列表前后分别加上 [0], 再利用zip 和sum 函数进行计算。 既解决了
# 两边都是1 的问题， 也造成原来的元素错位。sum 求和，就等于是两肩上的数求和！！ 厉害！并且只运用一层循环。
l1 = [sum(t) for t in zip([0]+l1, l1+[0])]
n += 1
return l2
print(generate(8))
Method 3: map method
- The same solution found on the Internet
def generate(numRows):
if numRows==0:
return []
l1 = [[1]]
n = 1
while n<numRows:
# 利用 map 函数 和匿名函数实现，本质和 zip num 实现是一样的，也是错位
l1.append(list(map(lambda x,y:x+y, [0]+l1[-1], l1[-1]+[0])))
n += 1
return l1
print(generate(8))