README.md

76. 最小覆盖子串

English Version

题目描述

给你一个字符串 s 、一个字符串 t 。返回 s 中涵盖 t 所有字符的最小子串。如果 s 中不存在涵盖 t 所有字符的子串，则返回空字符串 "" 。

 

注意：

• 对于 t 中重复字符，我们寻找的子字符串中该字符数量必须不少于 t 中该字符数量。
• 如果 s 中存在这样的子串，我们保证它是唯一的答案。

 

示例 1：

输入：s = "ADOBECODEBANC", t = "ABC"
输出："BANC"

示例 2：

输入：s = "a", t = "a"
输出："a"

示例 3:

输入: s = "a", t = "aa"
输出: ""
解释: t 中两个字符 'a' 均应包含在 s 的子串中，
因此没有符合条件的子字符串，返回空字符串。

 

提示：

• 1 <= s.length, t.length <= 105
• s 和 t 由英文字母组成

 

进阶：你能设计一个在 o(n) 时间内解决此问题的算法吗？

解法

滑动窗口

Python3

class Solution:
    def minWindow(self, s: str, t: str) -> str:
        ans = ''
        m, n = len(s), len(t)
        if m < n:
            return ans
        need = Counter(t)
        window = Counter()
        i, cnt, mi = 0, 0, float('inf')
        for j, c in enumerate(s):
            window[c] += 1
            if need[c] >= window[c]:
                cnt += 1
            while cnt == n:
                if j - i + 1 < mi:
                    mi = j - i + 1
                    ans = s[i: j + 1]
                c = s[i]
                if need[c] >= window[c]:
                    cnt -= 1
                window[c] -= 1
                i += 1
        return ans

Java

class Solution {
    public String minWindow(String s, String t) {
        Map<Character, Integer> mp = new HashMap<>();
        int begin = 0, end = 0, counter = t.length(), minLen = Integer.MAX_VALUE, minStart = 0, size = s.length();

        for (char c : s.toCharArray())
            mp.put(c, 0);
        for (char c : t.toCharArray()) {
            if (mp.containsKey(c))
                mp.put(c, mp.get(c) + 1);
            else
                return "";
        }

        while (end < size) {
            if (mp.get(s.charAt(end)) > 0)
                counter--;

            mp.put(s.charAt(end), mp.get(s.charAt(end)) - 1);

            end++;

            while (counter == 0) {
                if (end - begin < minLen) {
                    minStart = begin;
                    minLen = end - begin;
                }
                mp.put(s.charAt(begin), mp.get(s.charAt(begin)) + 1);

                if (mp.get(s.charAt(begin)) > 0) {
                    counter++;
                }

                begin++;
            }
        }

        if (minLen != Integer.MAX_VALUE) {
            return s.substring(minStart, minStart + minLen);
        }
        return "";
    }
}

TypeScript

function minWindow(s: string, t: string): string {
    let n1 = s.length,
        n2 = t.length;
    if (n1 < n2) return '';
    let need = new Array(128).fill(0);
    let window = new Array(128).fill(0);
    for (let i = 0; i < n2; ++i) {
        ++need[t.charCodeAt(i)];
    }

    let left = 0,
        right = 0;
    let res = '';
    let count = 0;
    let min = n1 + 1;
    while (right < n1) {
        let cur = s.charCodeAt(right);
        ++window[cur];
        if (need[cur] > 0 && need[cur] >= window[cur]) {
            ++count;
        }
        while (count == n2) {
            cur = s.charCodeAt(left);
            if (need[cur] > 0 && need[cur] >= window[cur]) {
                --count;
            }
            if (right - left + 1 < min) {
                min = right - left + 1;
                res = s.slice(left, right + 1);
            }
            --window[cur];
            ++left;
        }
        ++right;
    }
    return res;
}

C++

class Solution {
public:
    string minWindow(string s, string t){
        unordered_map<char, int> m;
        int begin = 0, end = 0, minlen = INT_MAX, minStart = 0, size = s.size(), counter = t.size();
        for (auto c: t)
            m[c]++;

        while (end < size) {
            if (m[s[end]] > 0)
                counter--;

            m[s[end]]--;

            end++;

            while (counter == 0) {
                if (end - begin < minlen) {
                    minStart = begin;
                    minlen = end - begin;
                }

                m[s[begin]]++;
                if (m[s[begin]] > 0)
                    counter++;

                begin++;
            }
        }

        if (minlen != INT_MAX) {
            return s.substr(minStart, minlen);
        }
        return "";
    }
};

Go

func minWindow(s string, t string) string {
	ans := ""
	m, n := len(s), len(t)
	if m < n {
		return ans
	}
	need := make([]int, 128)
	for _, c := range t {
		need[c] += 1
	}
	window := make([]int, 128)
	i, cnt, mi := 0, 0, m+1
	for j, c := range s {
		window[c]++
		if need[c] >= window[c] {
			cnt++
		}
		for cnt == n {
			if j-i+1 < mi {
				mi = j - i + 1
				ans = s[i : j+1]
			}
			c = rune(s[i])
			if need[c] >= window[c] {
				cnt--
			}
			window[c]--
			i++
		}
	}
	return ans
}

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