LeetCode Substring with Concatenation of All Words

最新推荐文章于 2019-12-26 08:25:23 发布

转载最新推荐文章于 2019-12-26 08:25:23 发布 · 47 阅读

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/lailailai/p/3617588.html

class Solution {
public:
    vector<int> findSubstring(string S, vector<string> &L) {
        vector<int> res;
		unordered_map<string, int> stat;
		unordered_map<string, int> run;
        int len = S.size();
		int num = L.size();
		int per = 0;

		if (num == 0 || !(per = L[0].size())) return res;
		
		int part= num * per;
		if (part > len) return res;
		
		int end = len - part;
        unordered_map<string, int>::iterator iter;
        pair<unordered_map<string, int>::iterator, bool> ir;
        
        for (int i=0; i<num; i++) {
			ir = stat.insert(pair<string, int>(L[i], 1));
			if (ir.second == false){
				ir.first->second++;				
			}
		}
		
		int i, j, pos, wc;
		string pre;
		for (i=0; i<=end; i++) {
			pos = i;
			for (j=0; j<num; j++, pos += per) {
				string seg = S.substr(pos, per);
				if (j == 0 || seg != pre) {
					iter = stat.find(seg);
					if (iter == stat.end()) break;
					wc = iter->second;
					ir = run.insert(pair<string, int>(seg, 1));
					iter = ir.first;
					if (ir.second) {
						pre = seg;
						continue;
					}
				}
				iter->second++;
				if (wc < iter->second) break;
			}
			if (j == num) res.push_back(i);
			run.clear();
    	}
    	return res;
    }
};

暴力解，跑了500+ms，如果每次都这样把题目应付过去，显然做题没有意义了。于是看Leetcode上关于那题的discuss，有人说可以用解决Minimum Window Substring的方法来解这一题，的确是可以。又搜了份Java实现的代码，一跑竟然也可以达到500ms，看来两种方法效率的确差很多。下面给出自己实现的代码

 1 class Solution3 {
 2 public:
 3     vector<int> findSubstring(string S, vector<string> &L) {
 4         vector<int> res;
 5         unordered_map<string, int> stat;
 6         unordered_map<string, int> run;
 7         int len = S.size();
 8         int num = L.size();
 9         int per = 0;
10 
11         if (num == 0 || !(per = L[0].size())) return res;
12         
13         int part = num * per;
14         if (part > len) return res;
15         
16         unordered_map<string, int>::iterator iter;
17         pair<unordered_map<string, int>::iterator, bool> ir;
18         
19         for (int i = 0; i < num; i++) {
20             ir = stat.insert(pair<string, int>(L[i], 0));
21             ir.first->second++;
22         }
23         int wc;
24         for (int i = 0; i < per; i++) {
25             int step = 0;
26             run.clear();
27             // scan like a worm, string[spos, epos] is the candidate
28             int spos=i, epos = i + per - 1;
29             for (; epos < len; epos += per) {
30                 string seg = S.substr(epos - per + 1, per);
31                 iter = stat.find(seg);
32                 // encounter some word not in L
33                 if (iter == stat.end()) {
34                     spos = epos + 1;
35                     step = 0;
36                     run.clear();
37                     continue;
38                 }
39                 
40                 wc = iter->second;
41                 step++;
42                 
43                 ir = run.insert(pair<string, int>(seg, 0));
44                 iter = ir.first;
45                 iter->second++;
46                 
47                 // string[spos, epos] is matched
48                 if (iter->second == wc && step == num) { 
49                     res.push_back(spos);
50                     run.find(S.substr(spos, per))->second--;
51                     step--;
52                     spos += per;
53                     continue;
54                 }
55                 
56                 // number of duplicated word exceeds needed
57                 if (iter->second > wc) {            
58                     string tmp = S.substr(spos, per);
59                     // find the first duplicated one
60                     while(seg != tmp) {
61                         run.find(tmp)->second--;
62                         step--;
63                         spos += per;
64                         tmp = S.substr(spos, per);
65                     }
66                     // then skip it
67                     iter->second--;
68                     spos += per;
69                     step--;
70                 }
71             }
72         }
73         return res;
74     } 
75 };

跑了一下在100+ms左右，又习得一策略技能！

参考：

zhuli题解 http://www.cnblogs.com/zhuli19901106/p/3570539.html

java实现 https://github.com/shengmin/coding-problem/blob/master/leetcode/substring-with-concatenation-of-all-words/Solution.java

leetcode资料 http://leetcode.com/2010/11/finding-minimum-window-in-s-which.html

转载于:https://www.cnblogs.com/lailailai/p/3617588.html