LeetCode 笔记系列七 Substring with Concatenation of All Words
2013-07-06 11:29
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题目: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:
L:
You should return the indices:
(order does not matter).
这道题想了蛮久,主要是coding过程中老是出一些边界的错误。第一眼看到,立(zi)马(zuo)想(cong)到(ming)的解法是不是可以用bit位来表示L中的各个key呀,不同的key代表不同的位,然后通过扫描S,通过与操作确定各个bit位是不是在长度为L的substring中。结果写了一会发现问题是,L中有可能有相等的字符串哦。。。那在搜索S的时候就悲剧了不是。
一个折中的方法是,对每个key在L中的,保存一个list,例如有L=["foo","bar","foo"],那我们做这样一个hash哇:
foo: 0->1
bar: 2
然后在搜索S的时候顺次检查一个bitset。代码如下:
解法一:
View Code
例如L=["foo","bar","foo"],我们的hash是这样的:
foo: 2
bar: 1
然后对S中每一个固定长度的substring做计数。
总结:
1.如果你用了超过3个(包括3个)复杂数据结构,应该思考是不是自己想多了;
2.想多了会[b]毁[/b]了你。
3.不要装b,先理解问题再想解法而不是为了用某个算法。
For example, given:
S:
"barfoothefoobarman"
L:
["foo", "bar"]
You should return the indices:
[0,9].
(order does not matter).
这道题想了蛮久,主要是coding过程中老是出一些边界的错误。第一眼看到,立(zi)马(zuo)想(cong)到(ming)的解法是不是可以用bit位来表示L中的各个key呀,不同的key代表不同的位,然后通过扫描S,通过与操作确定各个bit位是不是在长度为L的substring中。结果写了一会发现问题是,L中有可能有相等的字符串哦。。。那在搜索S的时候就悲剧了不是。
一个折中的方法是,对每个key在L中的,保存一个list,例如有L=["foo","bar","foo"],那我们做这样一个hash哇:
foo: 0->1
bar: 2
然后在搜索S的时候顺次检查一个bitset。代码如下:
解法一:
public static ArrayList<Integer> findSubstring3(String S, String[] L) { int lengthPerKey = L[0].length(); int numberOfKeys = L.length; HashMap<String, Integer> map = new HashMap<String, Integer>(); ArrayList<Integer> result = new ArrayList<Integer>(); for(String str : L){ if(!map.containsKey(str)){ map.put(str, 1); }else { map.put(str, map.get(str) + 1); } } for(int i = 0; i <= S.length() - numberOfKeys*lengthPerKey; i++){ int j = 0; int st = i; HashMap<String, Integer> wordsCount = new HashMap<String, Integer>(map); for(j = 0; j < numberOfKeys;j++){ if(S.length() - st < (numberOfKeys - j)*lengthPerKey) break; String sub = S.substring(st,st+lengthPerKey); if(wordsCount.containsKey(sub)){ int times = wordsCount.get(sub); if(times == 1) wordsCount.remove(sub); else wordsCount.put(sub, times - 1); }else break; st += lengthPerKey; } if(j == numberOfKeys){ result.add(i); } } return result; }
View Code
例如L=["foo","bar","foo"],我们的hash是这样的:
foo: 2
bar: 1
然后对S中每一个固定长度的substring做计数。
总结:
1.如果你用了超过3个(包括3个)复杂数据结构,应该思考是不是自己想多了;
2.想多了会[b]毁[/b]了你。
3.不要装b,先理解问题再想解法而不是为了用某个算法。
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