76 Minimum Window Substring

最新推荐文章于 2024-01-06 14:12:46 发布

dongbeier

最新推荐文章于 2024-01-06 14:12:46 发布

阅读量261

点赞数

CC 4.0 BY-SA版权

分类专栏： Linkedin LeetCode

本文链接：https://blog.youkuaiyun.com/dongbeier/article/details/80276140

LeetCode 同时被 2 个专栏收录

146 篇文章

订阅专栏

50 篇文章

订阅专栏

本文介绍了一种在字符串S中寻找包含字符串T所有字符的最短子串的算法，复杂度为O(n)。通过使用双指针技术，动态维护一个包含T所有字符的最小子串区间。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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).

Example:

Input: S = "ADOBECODEBANC", T = "ABC"
Output: "BANC"

Note:

If there is no such window in S that covers all characters in T, return the empty string "".
If there is such window, you are guaranteed that there will always be only one unique minimum window in S.

双指针，动态维护一个区间。尾指针不断往后扫，当扫到有一个窗口包含了所有 T 的字符后，然后再收缩头指针，直到不能再收缩为止。最后记录所有可能的情况中窗口最小的

class Solution {
    public String minWindow(String s, String t) {
        if (s.isEmpty()) {
            return "";
        }
        
        if (s.length() < t.length()) {
            return "";
        }
        final int ASCII = 256;
        int[] expect_count = new int[ASCII];
        int[] appear_count = new int[ASCII];
        Arrays.fill(expect_count, 0);
        Arrays.fill(appear_count, 0);
        
        int appear = 0, wnd_start = 0, min_width = Integer.MAX_VALUE;
        int min_start = 0;
        
        for (int i = 0; i < t.length(); i++) {
            expect_count[t.charAt(i)]++;
        }
        
        for (int wnd_end = 0; wnd_end < s.length(); wnd_end++) {
            if (expect_count[s.charAt(wnd_end)] > 0) {
                appear_count[s.charAt(wnd_end)]++;
                if (appear_count[s.charAt(wnd_end)] <= 
                   expect_count[s.charAt(wnd_end)]) {
                    appear++;
                }
            }
            
            if (appear == t.length()) {
                while (appear_count[s.charAt(wnd_start)] > 
                      expect_count[s.charAt(wnd_start)] || 
                       expect_count[s.charAt(wnd_start)] == 0) {
                    appear_count[s.charAt(wnd_start)]--;
                    wnd_start++;
                }
                if (min_width > wnd_end - wnd_start + 1) {
                    min_width = wnd_end - wnd_start + 1;
                    min_start = wnd_start;
                }
            }
        }
        if (min_width == Integer.MAX_VALUE) {
            return "";
        } else {
            return s.substring(min_start, min_start + min_width);
        }
    }
}