[leetcode]Minimum Window Substring @ Python

原題地址：https://oj.leetcode.com/problems/minimum-window-substring/

題意：

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 emtpy string "".

If there are multiple such windows, you are guaranteed that there will always be only one unique minimum window in S.

解題思路：雙指針思想，尾指針不斷往后掃，當掃到有一個窗口包含了所有T的字符，然后再收縮頭指針，直到不能再收縮為止。最后記錄所有可能的情況中窗口最小的。

代碼：

class Solution:
    # @return a string
    def minWindow(self, S, T):
        count1={}; count2={}
        for char in T:
            if char not in count1: count1[char]=1
            else: count1[char]+=1
        for char in T:
            if char not in count2: count2[char]=1
            else: count2[char]+=1
        count=len(T)
        start=0; minSize=100000; minStart=0
        for end in range(len(S)):
            if S[end] in count2 and count2[S[end]]>0:
                count1[S[end]]-=1
                if count1[S[end]]>=0:
                    count-=1
            if count==0:
                while True:
                    if S[start] in count2 and count2[S[start]]>0:
                        if count1[S[start]]<0:
                            count1[S[start]]+=1
                        else:
                            break
                    start+=1
                if minSize>end-start+1:
                    minSize=end-start+1; minStart=start
        if minSize==100000: return ''
        else:
            return S[minStart:minStart+minSize]