LeetCode 76. Minimum Window Substring

原题链接在这里: https://leetcode.com/problems/minimum-window-substring/

题目： 

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.

题解：

把t的char和对应frequency放到map中. count最开始是t.length()

[walker, runner]窗口. s在runner的char若在map中，对应frequency--. count--

当count == 0时, 窗口罩住所有t的char.

移动walker, 直到下一个s在walker的char, 若在map中, 对应frequency++. count++.

Time Complexity: O(sLen). sLen = s.length().

Space: O(1). 256数组map.

AC Java:

 1 public class Solution {
 2     public String minWindow(String s, String t) {
 3         if(s == null || t == null || s.length() < t.length()){
 4             return "";
 5         }
 6         
 7         int [] map = new int[256];
 8         for(char c : t.toCharArray()){
 9             map[c]++;
10         }
11         
12         int count = t.length();
13         int minLen = Integer.MAX_VALUE;
14         int head = 0;
15         int walker = 0; 
16         int runner = 0;
17         while(runner < s.length()){
18             if(map[s.charAt(runner++)]-- > 0){
19                 count--;
20             }
21             while(count == 0){  //count == 0时就表示罩住所有t的char
22                 if(runner - walker < minLen){
23                     minLen = runner - walker;
24                     head = walker;
25                 }
26                 if(map[s.charAt(walker++)]++ == 0){ //让其罩不住所有t的char
27                     count++;
28                 }
29             }
30         }
31         return minLen == Integer.MAX_VALUE ? "" : s.substring(head, head+minLen);
32     }
33 }

类似Minimum Size Subarray Sum, Longest Substring Without Repeating Characters, Longest Substring with At Most Two Distinct Characters, Substring with Concatenation of All Words.

对于快慢指针找substring这类问题，通用这个方法.

 1 int findSubstring(int s){
 2     int [] map = new int[256];
 3     int walker = 0;
 4     int runner = 0;
 5     int count; // check whether the substring is valid
 6     int len; // length of substring
 7 
 8     for(){
 9         /*
10         * initialize map.
11         */
12     }
13 
14     while(runner < s.length()){
15         if(map[s.charAt(runner++)]-- ?){
16             /*
17             * modify count.
18             */
19         }
20         while(/*count condition*/){
21 
22             // update len here if finding minimum
23             
24             if(map[s.charAt(walker++)]++ ?) {
25                 //move walker to make it invalid/valid again
26                 //modify count
27             }
28         }
29 
30         //update len here if finding maximum
31     }
32 
33     return len;
34 }

Note: 找maximum substring时，是在while loop 后面更新，因为需要while loop把substring 变valid. 

找minimum substring时，是在while loop之中找, 因为while loop是要把substring 变invalid.

好比Longest Substring Without Repeating Characters.

 1 public class Solution {
 2     public int lengthOfLongestSubstring(String s) {
 3         int res = 0;
 4         int [] map = new int[256];
 5         int count = 0;
 6         int walker = 0;
 7         int runner = 0;
 8         
 9         while(runner < s.length()){
10             if(map[s.charAt(runner++)]++ > 0){
11                 count++;
12             }
13             while(count > 0){
14                 if(map[s.charAt(walker++)]-- > 1){
15                     count--;
16                 }
17             }
18             res = Math.max(res, runner-walker);
19         }
20         return res;
21     }
22 }

和Longest Substring with At Most Two Distinct Characters

 1 public class Solution {
 2     public int lengthOfLongestSubstringTwoDistinct(String s) {
 3         int res = 0;
 4         int [] map = new int[256];
 5         int walker = 0;
 6         int runner = 0;
 7         int count = 0;
 8         while(runner < s.length()){
 9             if(map[s.charAt(runner++)]++ == 0){
10                 count++;
11             }
12             while(count > 2){
13                 if(map[s.charAt(walker++)]-- == 1){
14                     count--;
15                 }
16             }
17             res = Math.max(res, runner - walker);
18         }
19         return res;
20     }
21 }

都是在while loop后面找maximum的substring. 以为while loop 使 substring valid.

Reference: https://discuss.leetcode.com/topic/30941/here-is-a-10-line-template-that-can-solve-most-substring-problems

转载于:https://www.cnblogs.com/Dylan-Java-NYC/p/4825045.html