Pascal's Triangle II

最新推荐文章于 2022-02-07 21:31:57 发布

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最新推荐文章于 2022-02-07 21:31:57 发布

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分类专栏： leetcode

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本文提供了一种算法实现，用于根据给定的行索引返回帕斯卡三角形中对应的行元素，并优化了空间复杂度。

Given an index k, return the kth row of the Pascal’s triangle.

For example, given k = 3,
Return [1,3,3,1].

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

int* getRow(int rowIndex, int* returnSize) {

*returnSize = rowIndex + 1;

int** num = (int**)malloc(sizeof(int*) * (*returnSize));
for (int i = 0; i < *returnSize; i++)
    num[i] = (int*)malloc(sizeof(int) * (i+1));

for (int i = 0; i < *returnSize; i++)
{
    for (int j = 0; j <= i; j++)
    {
        if (j == 0 || j == i)
            num[i][j] = 1;
        else
            num[i][j] = num[i-1][j-1] + num[i-1][j];
    }
}

int* returnNum = (int*)malloc(sizeof(int)*(*returnSize));
for (int i = 0; i < *returnSize; i++)
    returnNum[i] = num[(*returnSize) - 1][i];


for (int i = 0; i < *returnSize; i++)
    free(num[i]);
free(num);

return returnNum;

}