(数组)leetcode 119:Pascal's Triangle II
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水平有限,不足之处还望指正。
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?
题意:
给定杨辉三角的当前行值,输出当前行,要求空间复杂度为O(K).
解题思路:
杨辉三角性很容易求,要求空间复杂度为O(K),表明只能在当前的数组上进行操作,不能开辟新空间。
操作当前数组,对于杨辉三角形,仅利用相邻的两个数据,很自然就想到从后向前遍历数组。
AC代码如下(含有main函数,VS2013中可直接运行):
#include <iostream>#include <vector>using namespace std;class Solution {public:vector<int> getRow(int rowIndex) {vector<int> result;if (0 > rowIndex)return result;for (int i = 0; i <= rowIndex; i++){for (int j = i-1; j >= 1; j--){result[j] = result[j] + result[j - 1];}result.push_back(1);}return result;}};int main(){vector<int> result;Solution my_solution;int rowIndex = 3;result = my_solution.getRow(rowIndex);for (int i = 0; i < result.size(); i++)cout << result[i] << " ";cout << endl;return 0;}
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