[LeetCode]Minimum Window Substring

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 empty string "".

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

分析

经典题型，类似于Longest Substring with At Most Two Distinct Characters。
此类window题型，我们都需要用两个指针，用一个map记录字符及其出现次数，不同的是由于这里题目要求是覆盖字符串T中所有字符，所以我们需要用一个变量如count来记录window中覆盖字符串T中有效字符的个数。

只要count == T.length(), 便可更新window的最短长度。同时，我们必须移动左指针，直到window中包含的字符个数小于规定的数量，我们才开始移动右指针。另外要注意何时更新count的值。

复杂度

time: O(n), space: O(n)

代码

public class Solution {
    public String minWindow(String s, String t) {
    
        // 建map, 记录被包含字符串中字符及其个数
        Map<Character, Integer> map = new HashMap<>();
        for (int i = 0; i < t.length(); i++) {
            char c = t.charAt(i);
            if (!map.containsKey(c)) {
                map.put(c, 1);
            } else {
                map.put(c, map.get(c) + 1);
            }
        }
        
        int l = 0;
        int minStart = 0;
        int minLen = s.length() + 1;
        int count = 0;
        
        for (int i = 0; i < s.length(); i++) {
            char c = s.charAt(i);
            if (map.containsKey(c)) {
                if (map.get(c) > 0)  
                    count++; // 注意count++的条件
                map.put(c, map.get(c) - 1);
            }
            while (count == t.length()) {
                
                // 不断更新最值
                if (i - l + 1 < minLen) {
                    minLen = i - l + 1;
                    minStart = l;
                }
                
                // 移动左指针
                char leftChar = s.charAt(l);
                if (map.containsKey(leftChar)) {
                    map.put(leftChar, map.get(leftChar) + 1);
                    if (map.get(leftChar) > 0) { 
                        count--; // 注意count--条件
                    }
                }
                l++;
            }
        }
        
        if (minLen == s.length() + 1) {
            return "";
        }
        return s.substring(minStart, minLen + minStart);
    }
}