LeetCode: Pascal's Triangle II [119]

class Solution {
public:
    vector<int> getRow(int rowIndex) {
        vector<int>result;
        if(rowIndex<0)return result;
        
        //维护两个rowIndex+1长度的数组, rowIndex为偶数时，行元素存储在row1中；rowIndex为奇数时，行元素存储在row2中
        vector<int> row1(rowIndex+1, 1);
        vector<int> row2(rowIndex+1, 1);
        int indexCount=0;   //当前已经生成的行的行索引
        while(indexCount<rowIndex){
            indexCount++;
            if(indexCount%2==0){
                //当前要从row2生成下一行元素，存到row1中
                //row2有indexCount个元素，row1中有indexCount+1个元素
                for(int i=1; i<indexCount; i++){
                    row1[i]=row2[i-1]+row2[i];
                }
            }
            else{
                //当前要从row1生成下一行元素，存到row2中
                //row1有indexCount个元素，row2中有indexCount+1个元素
                for(int i=1; i<indexCount; i++){
                    row2[i]=row1[i-1]+row1[i];
                }
            }
        }
        //如果rowIndex是偶数，返回row1，否则返回row2
        if(rowIndex%2==0)return row1;
        else return row2;
    }
};