【leetcode】118. Pascal's Triangle-优快云博客

本文介绍了一种简单高效的算法来生成帕斯卡三角。该算法的时间复杂度为O(n)，空间复杂度同样为O(n)。通过迭代计算每一层的数值，并利用前一层的数据来构造当前层，实现了对帕斯卡三角的快速生成。

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@requires_authorization
@author johnsondu
@create_time 2015.7.23 19:54
@url [Pascal's Triangle](https://leetcode.com/problems/pascals-triangle/)
/************************
 * @description: simple.
 * @time_complexity:  O(n)
 * @space_complexity: O(n)
 ************************/
 class Solution {
public:
    vector<vector<int>> generate(int numRows) {
        vector<vector<int>> ans;

        for(int i = 1; i <= numRows; i ++) {
            vector<int> layer;
            if(i == 1) layer.push_back(1);
            else {
                for(int j = 1; j <= i; j ++) {
                    if(j == 1 || j == i)  layer.push_back(1);
                    else layer.push_back(ans[i-2][j-2] + ans[i-2][j-1]);
                }
            }
            ans.push_back(layer);
        }
        return ans;
    }
};

@requires_authorization
@author johnsondu
@create_time 2015.7.23 19:54
@url [Pascal's Triangle](https://leetcode.com/problems/pascals-triangle/)
/************************
 * @description: simple.
 * @time_complexity:  O(n)
 * @space_complexity: O(n)
 ************************/
class Solution {
public:
    vector<vector<int>> generate(int numRows) {
        vector<vector<int>> ans;
        if(numRows < 1) return ans;

        vector<int> first;
        first.push_back(1);
        ans.push_back(first);
        if(numRows < 2) return ans;

        vector<int> second;
        second.push_back(1);
        second.push_back(1);
        ans.push_back(second);

        for(int i = 3; i <= numRows; i ++) {
            vector<int> layer;
            layer.push_back(1);
            for(int j = 0; j < ans[i-2].size()-1; j ++) {
                layer.push_back(ans[i-2][j] + ans[i-2][j+1]);
            }
            layer.push_back(1);
            ans.push_back(layer);
        }
        return ans;
    }
};