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LeetCode: pascal‘s Triangle
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]
]
地址：https://oj.leetcode.com/problems/pascals-triangle/
算法：pascal‘s triangle的特点是第i+1行第j个元素为第i行第j个元素和第j-1个元素的和，当然头尾两个元素要特殊处理。掌握这个特点，写出代码应该不难：
1 class Solution {
2 public:
3 vector<vector<int> > generate(int numRows) {
4 vector<vector<int> > result;
5 if(numRows < 1){
6 return result;
7 }
8 result.push_back(vector<int>(1,1));
9 for(int i = 1; i < numRows; ++i){
10 vector<int> temp(i+1);
11 temp[0] = 1;
12 temp[i] = 1;
13 for(int j=1; j < i; ++j){
14 temp[j] = result[i-1][j] + result[i-1][j-1];
15 }
16 result.push_back(temp);
17 }
18 return result;
19 }
20 };
第二题：
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?
地址：https://oj.leetcode.com/problems/pascals-triangle-ii/
算法：要产生第k行的数组必须先产生第k-1行数组，但跟第k-2行的数组完全无关，所以可以用上一篇文章的方法来达到空间上的限制，而产生数组的方式跟第一题的方法一样。代码：
1 class Solution {
2 public:
3 vector<int> getRow(int rowIndex) {
4 if(rowIndex < 0) return vector<int>();
5 if(rowIndex == 0) return vector<int>(1,1);
6 vector<int> result1(rowIndex+1);
7 vector<int> result2(rowIndex+1);
8 result1[0] = 1;
9 vector<int> *pre = &result1;
10 vector<int> *p = &result2;
11 for(int i = 1; i <= rowIndex; ++i){
12 (*p)[0] = 1;
13 (*p)[i] = 1;
14 for(int j = 1; j < i; ++j){
15 (*p)[j] = (*pre)[j] + (*pre)[j-1];
16 }
17 vector<int> *temp = pre;
18 pre = p;
19 p = temp;
20 }
21 return *pre;
22 }
23 };
标签：style blog http color os io strong for ar
原文地址：http://www.cnblogs.com/boostable/p/leetcode_pascal_triangle.html
