LeetCode OJ：Pascal's Triangle II

题目：

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?

思路：

因为每一层的数都是对称的，所以只需要计算每一层的前半部分的值，后半部拷贝就可以了。res[i] = res`[i - 1] + res`[i]（res`代表原数组，res代表更新后的数组），tmp代表res`[i - 1]，tmp2代表 res`[i]。

代码：

int* getRow(int rowIndex, int* returnSize) {
    if(rowIndex < 0) return NULL;
    ++rowIndex;
    int *res = (int*)malloc(sizeof(int) * rowIndex);
    res[0] = 1;
    int count = rowIndex;
    while(--count){
        int real_index = rowIndex - count + 1;
        int half = real_index / 2;
        if(real_index % 2 == 0){
            half--;
        }
        int tmp = res[0];
        int i = 1;
        for(; i <= half; i++){
            int tmp2 = res[i];
            res[i] = tmp + tmp2;
            tmp = tmp2;
        }
        if(real_index % 2){
            i -= 2;
        } else {
            --i;
        }
        for(int j = half + 1; j < real_index; j++){
            res[j] = res[i--];
        }
    }
    *returnSize = rowIndex;
    return res;
}

LeetCode OJ：Pascal's Triangle II