【leetcode 76】 Minimum Window Substring
2017-03-01 19:52
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/*********************************************************************
Given a string S and a string T, find the minimum window in S which will contain all the characters in T in complexity O(n).
For example,
S =
T =
Minimum window is
Note:
If there is no such window in S that covers all characters in T, return the empty string
If there are multiple such windows, you are guaranteed that there will always be only one unique minimum window in S.
/***********************************************************************
思路:
1,用 双指针法,两个指针夹的子字符串就是 Minimum window。【在字符串或数组中扣出指定的子字符串或子数组常常用这种思想】
2,用一个map来储存字符和出现次数的关系,map<char,int>。
3,初始化map时,用T中字符出现的字符次数来初始化,如:T="aabc",则 map[a]=2,map[b]=1,map[c]=1;
4,遍历 S, S中字符x每出现一次,map[x]--;用 count 来统计T在S中还需出现的数目。初始化 count=t.size();
每出现一次T中字符 count--;当count==0时表示S出现完了T中字符。
/**********************************************************************
代码:
class Solution{
public:
string minWindow(string s, string t){
map<char, int> map; //(128, 0);
for (auto c : t) map[c]++;
int count = t.size(), begin = 0, end = 0, head = 0,d=INT_MAX;
while (end < s.size()){
if (map[s[end++]]-->0) count--;
while (count == 0){
if (d > (end - begin)){
d = end - begin;
head = begin;
}
if (map[s[begin++]]++ == 0) count++; // 为什么是 map[s[begin]]++ ??因为往后移动了就还缺一个 s[begin].
}
}
return d == INT_MAX ? "" : s.substr(head,d);
}
};
Given a string S and a string T, find the minimum window in S which will contain all the characters in T in complexity O(n).
For example,
S =
"ADOBECODEBANC"
T =
"ABC"
Minimum window is
"BANC".
Note:
If there is no such window in S that covers all characters in T, return the empty string
"".
If there are multiple such windows, you are guaranteed that there will always be only one unique minimum window in S.
/***********************************************************************
思路:
1,用 双指针法,两个指针夹的子字符串就是 Minimum window。【在字符串或数组中扣出指定的子字符串或子数组常常用这种思想】
2,用一个map来储存字符和出现次数的关系,map<char,int>。
3,初始化map时,用T中字符出现的字符次数来初始化,如:T="aabc",则 map[a]=2,map[b]=1,map[c]=1;
4,遍历 S, S中字符x每出现一次,map[x]--;用 count 来统计T在S中还需出现的数目。初始化 count=t.size();
每出现一次T中字符 count--;当count==0时表示S出现完了T中字符。
/**********************************************************************
代码:
class Solution{
public:
string minWindow(string s, string t){
map<char, int> map; //(128, 0);
for (auto c : t) map[c]++;
int count = t.size(), begin = 0, end = 0, head = 0,d=INT_MAX;
while (end < s.size()){
if (map[s[end++]]-->0) count--;
while (count == 0){
if (d > (end - begin)){
d = end - begin;
head = begin;
}
if (map[s[begin++]]++ == 0) count++; // 为什么是 map[s[begin]]++ ??因为往后移动了就还缺一个 s[begin].
}
}
return d == INT_MAX ? "" : s.substr(head,d);
}
};
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