【Pascal's Triangle】cpp

题目:

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]
]

代码:

class Solution {
public:
    vector<vector<int>> generate(int numRows) {
            vector<vector<int> > ret;
            if ( numRows<1 ) return ret;
            vector<int> pre, curr;
            for ( int i=0; i<numRows; ++i )
            {
                curr.clear();
                curr.push_back(1);
                for ( int j=0; j<pre.size(); ++j )
                {    
                    if ( j==pre.size()-1 ){
                        curr.push_back(1);
                    }
                    else{
                        curr.push_back(pre[j]+pre[j+1]);
                    }
                }
                pre = curr;
                ret.push_back(curr);
            }
            return ret;
    }
};

tips:

数组基本操作。

==============================================

第二次过这道题,代码更简洁了。

class Solution {
public:
    vector<vector<int>> generate(int numRows) {
            vector<vector<int> > ret;
            if ( numRows<1 ) return ret;
            vector<int> pre;
            pre.push_back(1);
            ret.push_back(pre);
            for ( int i=1; i<numRows; ++i )
            {
                pre = ret.back();
                vector<int> curr;
                curr.push_back(1);
                for ( int j=1; j<pre.size(); ++j ) curr.push_back(pre[j-1]+pre[j]);
                curr.push_back(1);
                ret.push_back(curr);
            }
            return ret;
    }
};

===========================================

第三版,更简洁了一些

class Solution {
public:
    vector<vector<int>> generate(int numRows) {
        vector<vector<int> > ret;
        if (numRows==0) return ret;
        vector<int> tmp;
        tmp.push_back(1);
        ret.push_back(tmp);
        for ( int i=1; i<numRows; ++i )
        {
            for ( int j=tmp.size(); j>0; --j ) tmp[j] = tmp[j] + tmp[j-1];
            tmp.push_back(1);
            ret.push_back(tmp);
        }
        return ret;
    }
};

posted on 2015-06-08 23:14  承续缘  阅读(176)  评论(0编辑  收藏  举报

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