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?

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Hide Tags ?Array

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方法一：保存所有二位數(shù)組

class Solution {public:vector<int> getRow(int rowIdx) {vector<int> curLine;vector<vector<int> > res;curLine.push_back(1);res.push_back(curLine);if( rowIdx == 0)return curLine;for(int i = 1; i <= rowIdx; i++){curLine.clear();for(int j = 0; j < res[i-1].size(); j++){if(j == 0)curLine.push_back(1);elsecurLine.push_back(res[i-1][j-1] + res[i-1][j]);}curLine.push_back(1);res.push_back(curLine);}return res[rowIdx];} };

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方法二：cur只和上一行有關(guān)，用滾動數(shù)組即可實現(xiàn)空間復(fù)雜度O（n）

class Solution {public:vector<int> getRow(int rowIdx) {vector<int> preLine;vector<int> curLine;curLine.push_back(1);if(rowIdx == 0)return curLine;for(int i = 1; i <= rowIdx; i++){preLine = curLine;curLine.clear();for(int j = 0; j < preLine.size(); j++){if(j == 0)curLine.push_back(1);elsecurLine.push_back(preLine[j-1] + preLine[j]);}curLine.push_back(1);}return curLine;} };

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轉(zhuǎn)載于:https://www.cnblogs.com/diegodu/p/4421725.html

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