[LeetCode]ZigZag Conversion

最新推荐文章于 2021-11-05 22:58:08 发布

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最新推荐文章于 2021-11-05 22:58:08 发布

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分类专栏： LeetCode

本文链接：https://blog.youkuaiyun.com/sunbaigui/article/details/8981384

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本文提供了一种将字符串按Z形排列进行转换的方法，并通过两种不同的C++实现方案进行了详细说明。一种方案利用了二维向量来模拟Z形排列的过程，另一种方案则采用数学计算的方式直接确定每个字符在转换后的字符串中的位置。

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class Solution {
//be careful with the special case
//when done with coding, figure out some cases (including illegal, normal, edge cases)
//to run with this solution, this will help to get a bug-free solution
public:
	string convert(string s, int nRows) {
		// Start typing your C/C++ solution below
		// DO NOT write int main() function    
		vector<vector<char>> zig(nRows, vector<char>());
		int zigRow = 0;
		int sign = 1;
		for (int i = 0; i < s.size(); ++i)
		{
			zig[zigRow].push_back(s[i]);
			if (nRows == 1)//special case
				zigRow = 0;
			else
			{
				if(zigRow == nRows-1) sign = -1;
				if(zigRow == 0) sign = 1;
				zigRow += sign;
			}
			
		}
		string ans;
		for (int i = 0; i < zig.size(); ++i)
		{
			for (int j = 0; j < zig[i].size(); ++j)
			{
				ans.push_back(zig[i][j]);
			}
		}
		return ans;
	}
};

second time

class Solution {
public:
    string convert(string s, int nRows) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function    
        if(nRows == 1) return s;
        
        string result;
        for(int i = 0; i < nRows; ++i)
        {
            int curOriginalIdx = i;
            int k = 0;
            while(curOriginalIdx < s.size())
            {
                int nextOriginalIdx;
                if(k%2 == 0) nextOriginalIdx = curOriginalIdx+2*(nRows-1-i);
                else nextOriginalIdx = curOriginalIdx+2*(i);
                if(nextOriginalIdx != curOriginalIdx)
                    result.push_back(s[curOriginalIdx]);
                curOriginalIdx = nextOriginalIdx;
                k++;
            }
        }
        
        return result;
    }
};