LeetCode 119. Pascal's Triangle II

题目

Given a non-negative index k where k ≤ 33, return the k^th index row of the Pascal's triangle.

Note that the row index starts from 0.

In Pascal's triangle, each number is the sum of the two numbers directly above it.
Example:

Input: 3
Output: [1,3,3,1]

Follow up:
Could you optimize your algorithm to use only O(k) extra space?

解析

典型的杨辉三角求某一行的所有值。此处要注意的是rowIndex从0开始，题目中要求优化空间复杂度为O(k)，这意图就很明显了，即让我们找到规律后依次赋值。

二项展开式

右侧的部分也可以写成，
n(n-1)(n-2)…(n-r+1) / r!
写成这种形式以便在求组成数进行循环的时候比较方便。

代码

// 求组合数，使用long以防止int越界
long getCombination(int n, int m) {
    long res = 1;
    
    for (int i = 1; i <= m; ++i) {
        res = (res * (n - m + i)) / i;
    }
    
    return res;
}

int* getRow(int rowIndex, int* returnSize) {
    (* returnSize) = rowIndex + 1;
    int* returnArr = (int*)malloc((*returnSize) * sizeof(int));
    
    returnArr[0] = returnArr[(*returnSize) - 1] = 1;
    
    if (rowIndex == 0) {
        returnArr[0] = 1;
        
        return returnArr;
    }
    
    for (int i = 1; i < ((*returnSize) / 2 + 1); ++i) {
        returnArr[i] = returnArr[(*returnSize) - i - 1] =
        (int)getCombination(rowIndex, i);     // 强转为int
    }
    
    return returnArr;
}