leetcode第29题--Substring with Concatenation of All Words
problem:
You are given a string, S, and a list of words, L, that are all of the same length. Find all starting indices of substring(s) in S that is a concatenation of each word in L exactly once and without any intervening characters.
For example, given:
S: "barfoothefoobarman"
L: ["foo", "bar"]
You should return the indices: [0,9]
.
(order does not matter).
题目的意思是,给定S和字符串数组L,判断S中是否存在包含L中的所有字符串,字符串的顺序不做要求,但是字符串之间不得有其他字符,如果存在,返回S中的下标。例如“abcba”,L为“a”,“b”,那么返回0和3. 还有就是允许之间的重复,也就是“aba”,L为“a”,“b”时,返回为0和1,其中b被重复考虑。
做法是这样的,先创建两个map<string, int> 一个是用来计算L中每种串的出现次数。另外一个是用来在遍历过程中判断超出了第一个map表记录的L中的次数。第二个map记得每次重新遍历的时候要清空。详细代码如下:
class Solution {
public:
vector<int> findSubstring(string S, vector<string> &L)
{
vector<int> ans;
if (S.size() == || L.size() == || S.size() < L.size() * L[].size())
return ans;
int wordNum = L.size();
int wordLen = L[].size();
int searchEnd = S.size() - wordNum * wordLen;
map<string, int> total;
map<string, int> subMap; for(int i = ; i < wordNum; i++)
{
total[L[i]]++;
} for (int i = ; i <= searchEnd; i++) // 等号不能少了,还有最后就是用searchEnd代替S.size() - wordNum*wordLen;
{
int matchNum = , j = i;
subMap.clear();// 记得清空
for (; matchNum < wordNum; matchNum++)
{
string subs = S.substr(j, wordLen);
if (total[subs] == )
break;
if (++subMap[subs] > total[subs])
break;
j += wordLen;
}
if (matchNum == wordNum)
ans.push_back(i);
}
return ans;
}
};
还可以参见http://www.tuicool.com/articles/Uza2eui;java的参见http://blog.csdn.net/linhuanmars/article/details/20342851这个方法比较快,有时间一定要学学。
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