LeetCode OJ - Substring with Concatenation of All Words

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).

分析：首先采用最普通的匹配方法，匹配的过程中注意，是要匹配完所有L中的单词。L中的匹配用循环+hash计数来统计。

/* i from S   j in L      1.当j在L中匹配玩，记录      2.没匹配完，不记录*/class Solution {    vector<int> ret;public:    vector<int> findSubstring(string S, vector<string> &L) {        int num = L.size();        int len = L[0].size();        //L中单词出现的次数        map<string, int> words;        for(int i = 0; i < num; i++) {            words[L[i]]++;        }                map<string, int> tmp;        for(int i = 0; i <= S.size() - num * len; i++) {            tmp.clear();            int j;            for(j = 0; j < num; j++) {                string word = S.substr(i+j*len, len);                if(words.find(word) == words.end())                    break;                else                    tmp[word]++;                if(tmp[word] > words[word])                    break;            }                        if(j == num)                ret.push_back(i);        }                return ret;    }};

这里的代码是错误的，因为string.size()返回的是无符号数。若无符号数与有符号数作减法，是很容易溢出的。

例如：unsigned int (1) - signed int (2) 

结果：若结果要转化为有符号数， 2^32 - 1; 若结果是无符号数，那么结果就是 - 1

原因：无符号数和有符号数运算时，都会被转化为无符号数

class Solution {    vector<int> ret;public:    vector<int> findSubstring(string S, vector<string> &L) {int num = L.size();int width = L[0].size();int lenS = S.size();map<string, int> hash;for(int i = 0; i < num; i++) {    hash[L[i]]++;}map<string, int> item;for(int i = 0; i <= lenS - num * width; i++) {    item.clear();    int j;    for(j = 0; j < num; j++) {        string word = S.substr(i + j * width, width);        if(hash.find(word) == hash.end()) break;        else item[word]++;                if(item[word] > hash[word]) break;    }        if(j == num) ret.push_back(i); }return ret;}};