【Medium*2】 Word Break 、ZigZag Conversion - leetCode

139

Word Break

34.3%

Medium

139. Word Break

Medium

1867107FavoriteShare

Given a non-empty string s and a dictionary wordDict containing a list of non-empty words, determine if s can be segmented into a space-separated sequence of one or more dictionary words.

Note:

• The same word in the dictionary may be reused multiple times in the segmentation.
• You may assume the dictionary does not contain duplicate words.

Example 1:

Input: s = "leetcode", wordDict = ["leet", "code"]
Output: true
Explanation: Return true because "leetcode" can be segmented as "leet code".

Example 2:

Input: s = "applepenapple", wordDict = ["apple", "pen"]
Output: true
Explanation: Return true because "applepenapple" can be segmented as "apple pen apple".
             Note that you are allowed to reuse a dictionary word.

Example 3:

Input: s = "catsandog", wordDict = ["cats", "dog", "sand", "and", "cat"]
Output: false

Accepted

303,378

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Submissions

883,755

DP+暴力

class Solution {
public:
    bool dp[100001];
    bool wordBreak(string s, vector<string>& wordDict) {
        //dp[i]=可以由dict拼接出前i个字符
        s="0"+s;
        string t;
        memset(dp,0,sizeof(dp));
        dp[0]=1;
        for(int i=0;i<=s.length();i++) //n
        {
            
            if(dp[i]==0)
                continue;
            for(int j=0;j<wordDict.size();j++) // size
            {
                if(s.substr(i+1,wordDict[j].length())==wordDict[j]) // m
                {
                    dp[i+wordDict[j].length()]=1;
                }
            }
        }
        return dp[s.length()-1];
    }
};

6

ZigZag Conversion

30.7%

Medium

6. ZigZag Conversion

Medium

9032832FavoriteShare

The string "PAYPALISHIRING" is written in a zigzag pattern on a given number of rows like this: (you may want to display this pattern in a fixed font for better legibility)

P   A   H   N
A P L S I I G
Y   I   R

And then read line by line: "PAHNAPLSIIGYIR"

Write the code that will take a string and make this conversion given a number of rows:

string convert(string s, int numRows);

Example 1:

Input: s = "PAYPALISHIRING", numRows = 3
Output: "PAHNAPLSIIGYIR"

Example 2:

Input: s = "PAYPALISHIRING", numRows = 4
Output: "PINALSIGYAHRPI"
Explanation:

P     I    N
A   L S  I G
Y A   H R
P     I

Accepted

288,258

Submissions

939,295

最容易想到的：使用vector或string压缩列间距即可，注意numRows==1特判。后面还有更优的方法。

class Solution {
public:
    string convert(string s, int numRows) {
        if(numRows==1)
        {
            return s; //这里很重要
        }
        vector<char>mp[1000];
        for(int i=0;i<=numRows;i++)
            mp[i].clear();
        int y,d;
        y=0;
        d=1;
        for(int i=0;i<s.length();i++)
        {
            mp[y].push_back(s[i]);
            y+=d;
            if(y>=numRows)
            {
                y-=2;
                d=-1;
            }
            else if(y<0)
            {
                y+=2;
                d=1;
            }
        }
        string ret="";
        for(int i=0;i<numRows;i++)
        {
            for(int j=0;j<mp[i].size();j++)
            {
                ret=ret+mp[i][j];
            }
        }
        return ret;
    }
};

当然，，更优的方法就是找规律，然后省略掉模拟的过程和空间。(就是麻烦点)

class Solution {
public:
    string convert(string s, int numRows) {
        if(numRows==1)
        {
            return s; //这里很重要
        }
        int cicle=numRows*2-2;
        string ret("");
        for(int i=0;i<s.length();) // 第一行
        {
            ret=ret+s[i];
            i+=cicle;
        }
        //中间
        int idx[2];
        int t;
        for(int i=1;i<=numRows-2;i++)
        {
            idx[0]=i;
            idx[1]=numRows-1+(numRows-1-i);
            for(int j=0;;j++)
            {
                t=j/2*cicle+idx[j&1];
                if(t>=s.length())
                    break;
                ret=ret+s[t]; 
            }
        }
        
        for(int i=numRows-1;i<s.length();) //最后一行
        {
            ret=ret+s[i];
            i+=cicle;
        }
        return ret;
    }
};