LeetCode | Minimum Window Substring

 

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.

//字符串t是可能包含重复字符的,ABCC的前三个元素就能完全包含ABC的字符,故判断窗口何时已完全包含t的字符是个难点
//思路是用双指针维护一个窗口,当窗口完全包含的t的字符时,移动head直到不能再移动为止
public class Solution {
    public String minWindow(String s, String t) {
        
        if(s==null || s.length()==0) {
            return "";
        }
        
        //根据字符串t建立一个字典:map< char, char在t中重复的次数 >
        Map<Character, Integer> map = new HashMap<Character, Integer>();
        for(int i=0; i<t.length(); i++){
            char myChar = t.charAt(i);
            if(map.containsKey(myChar)){
                map.put(myChar, map.get(myChar)+1);
            }else{
                map.put(myChar, 1);
            }
        }
        
        int head = 0;                                   //头指针
        int count = 0;                                  //用来判断窗口是否已经完全包含t的字符
        int start = 0;                                  //最小窗口的起始位置
        int minLength = Integer.MAX_VALUE;              //最小窗口的长度
        
        for(int tail=0; tail<s.length(); tail++){       //尾指针不断向后遍历
            char tailChar = s.charAt(tail);
            if(map.containsKey(tailChar)){              //如果不包含,就直接continue
                map.put(tailChar, map.get(tailChar)-1);
                if(map.get(tailChar) >= 0){       //在串t中仅出现了一次的char,仅仅能让count加1
                    count++;                      //当count加到t.length时,窗口[head-tail]一定完整的包含t串了,只可能多,不可能少
                }                                 //若t=ABCC,当tail遍历到窗口ADOBECC的第二个C,此时count=4,但第二个C是多余的
                
                while(count == t.length()){       //此时窗口[head-tail]已经完整包含t了,向右移动head,直到
                    if(tail-head+1 < minLength){  //窗口不在包含t了
                        minLength = tail-head+1;  //当前的窗口比以前选定的窗口更小,就更新最小窗口
                        start = head;
                    }
                    char headChar = s.charAt(head);
                    if(map.containsKey(headChar)){              //窗口head处的字符位于串t之中
                        map.put(headChar, map.get(headChar)+1);
                        if(map.get(headChar) > 0){              //说明在上面加一操作之前,map.get(headChar)=0
                            count--;                            //说明headChar是必须的,不能少了的
                        }
                    }
                    head++;
                }
            }
        }
        
        if(minLength > s.length()){
            return "";
        }
        
        return s.substring(start, start+minLength);
    }
}



posted @ 2015-06-29 20:44  Mr.do  阅读(99)  评论(0编辑  收藏  举报