描述

Given numRows, generate the first numRows of Pascal's triangle.

For example, given numRows = 5,

Return

[

[1],

[1,1],

[1,2,1],

[1,3,3,1],

[1,4,6,4,1]

]

分析

其实就是杨辉三角，以前用队列写过。

为了和numRows相匹配，以变量i代表当前的行数，那么i-1才是当前行的下标。以j代表当前行的第i个元素（这个j是从下标1开始的）。

代码如下：

class Solution {

public:

    vector<vector<int>> generate(int numRows) {

        vector<int>ele; //store elements of current line

        vector<vector<int>>ret; //store all the lines and as a return variable

        if(numRows == 0)return ret;     //return an empty vector if numRows == 0

        int j = 0;  //initialize j

        for(int i = 1; i <= numRows; ++i){  //note: i begins from 1,means the first line

            ele.push_back(1);   //push 1 before do any operators in a line

            if(i >= 2){

                j = 2;

                while(j < i){

                    //because i starts with 1,so i - 1 is current line,while i - 2 is the line before current

                    //so as to j

                    ele.push_back(ret[i - 2][j - 2] + ret[i - 2][j - 1]);

                    ++j;

                }

                ele.push_back(1);   //the last number in a line is also 1

            }

            ret.push_back(ele);

            ele.clear();    //each time we finish a line,clear this vector

        }

        return ret;

    }

};

巴特西

leetcode解题报告（23）：Pascal's Triangle

描述

分析

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