Leecode刷题之旅-C语言/python-118杨辉三角

/*
 * @lc app=leetcode.cn id=118 lang=c
 *
 * [118] 杨辉三角
 *
 * https://leetcode-cn.com/problems/pascals-triangle/description/
 *
 * algorithms
 * Easy (60.22%)
 * Total Accepted:    17.6K
 * Total Submissions: 29.2K
 * Testcase Example:  '5'
 *
 * 给定一个非负整数 numRows，生成杨辉三角的前 numRows 行。
 *
 *
 *
 * 在杨辉三角中，每个数是它左上方和右上方的数的和。
 *
 * 示例:
 *
 * 输入: 5
 * 输出:
 * [
 * ⁠    [1],
 * ⁠   [1,1],                                    0
 * ⁠  [1,2,1],
 * ⁠ [1,3,3,1],
 * ⁠[1,4,6,4,1]
 * ]
 *
 */
/**
 * Return an array of arrays.
 * The sizes of the arrays are returned as *columnSizes array.
 * Note: Both returned array and *columnSizes array must be malloced, assume caller calls free().
 */
int** generate(int numRows, int** columnSizes) {
    int i,j;
    if(numRows == ) return ;
    int** Array = (int **)malloc(numRows * sizeof(int *));
    *columnSizes = (int *)malloc(numRows * sizeof(int));
    for(i = ; i < numRows; i++){
        (*columnSizes)[i] = i + ;
         Array[i] = (int *)malloc((i + ) * sizeof(int));
        for(j = ; j < i + ; j++){
          if((j == ) || (j == i))
          Array[i][j] = ;
           else
           Array[i][j] = Array[i - ][j - ] + Array[i - ][j];
        }
     }
 return Array;
}

算法核心是很好理解的。如果是首位或者末位，就等于一。否则的话，等于上一轮中两数之和。 如果当前是a[i][j] 那么就等于 a[i-1][j]+a[i-1][j+1]

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python：

#
# @lc app=leetcode.cn id=118 lang=python3
#
# [118] 杨辉三角
#
# https://leetcode-cn.com/problems/pascals-triangle/description/
#
# algorithms
# Easy (60.22%)
# Total Accepted:    17.6K
# Total Submissions: 29.2K
# Testcase Example:  '5'
#
# 给定一个非负整数 numRows，生成杨辉三角的前 numRows 行。
#
#
#
# 在杨辉三角中，每个数是它左上方和右上方的数的和。
#
# 示例:
#
# 输入: 5
# 输出:
# [
# ⁠    [1],
# ⁠   [1,1],
# ⁠  [1,2,1],
# ⁠ [1,3,3,1],
# ⁠[1,4,6,4,1]
# ]
#
#
class Solution:
    def generate(self, numRows):
        result = []
        if numRows == 0: return []
        for i in range(numRows):
            temp = []
            for j in range(i + 1):
                if j == 0:
                    temp.append(1)
                elif j == i:
                    temp.append(1)
                else:
                    add = result[i - 1][j - 1] + result[i - 1][j]
                    temp.append(add)
            result.append(temp)
        return result