LintCode：M-单词切分

LintCode链接

给出一个字符串s和一个词典，判断字符串s是否可以被空格切分成一个或多个出现在字典中的单词。

您在真实的面试中是否遇到过这个题？ 

Yes

样例

给出

s = "lintcode"

dict = ["lint","code"]

返回 true 因为"lintcode"可以被空格切分成"lint code"

public class Solution {    /*     * @param s: A string     * @param dict: A dictionary of words dict     * @return: A boolean     */     //方法一         //AC=100%    public boolean wordBreak(String s, Set<String> dict) {        if(s.length() == 0 && (dict == null || dict.size()==0)) return true;        if(s.length() == 0 || (dict == null || dict.size()==0)) return false;                //boolen比int占的空间小        boolean[] dp = new boolean[s.length()+1];        dp[0] = true;        for(int i = 0; i < s.length(); i++){            for(String str : dict){                if(dp[i] && i+str.length() <= s.length() && s.substring(i,i+str.length()).equals(str)) dp[i+str.length()] = true;            }        }        return dp[s.length()];    }    //方法二        //AC=92%    //超内存    //栈空间不够用了，递归次数太多    public boolean wordBreak_1(String s, Set<String> dict) {                int minLen=Integer.MAX_VALUE, maxLen=Integer.MIN_VALUE;        for(String sTmp:dict){            minLen = Math.min(minLen, sTmp.length());            maxLen = Math.max(maxLen, sTmp.length());        }                    if(s.length()==0 && dict.size()==0)            return true;        if(s.length()==0 || dict.size()==0)            return false;                return helper(s, dict, minLen, maxLen);    }     String sub;    boolean helper(String s, Set<String> dict, int minLen, int maxLen){        for(int i=minLen; i<=maxLen; i++){            if(s.length()<i)                return false;            sub = s.substring(0, i);            if(dict.contains(sub)){                if(i==s.length())                    return true;                if(helper(s.substring(i, s.length()), dict, minLen, maxLen))                    return true;            }                        }        return false;    }    //方法三          //AC=96%    //超内存    //SC = O(n)    public boolean wordBreak_2(String s, Set<String> dict) {        int n=s.length();        if(n==0 && dict.size()==0) return true;                //dp[i] = 长i的字符串能够切分        int[] dp = new int[n+1];        dp[0]=1;        for(int i=1; i<=n; i++){//长 i-1 的串            for(int j=i-1; j>=0; j--){                String sub = s.substring(j,i);                if(dp[j]==1 && dict.contains(sub)){                    dp[i]=1;                    break;                }            }        }                return dp[n]==1?true:false;        }}