Problem: 76. 最小覆盖子串

题目描述

思路

1.定义两个map集合need和window（以字符作为键，对应字符出现的个数作为值），将子串t存入need中；
2.定义左右指针left、right均指向0（形成窗口），定义int类型变量len记录最小窗口长度,valid记录当前窗口否存最短子串中的字符个数；
3.向右扩大窗口，遍历到到的字符c如果在need中时，window[c]++，同时如果window[c] == need[c]，则valid++；
4.如果valid == need.size()，则表示可以开始收缩窗口，并更新最小窗口禅读（如果移除的字符在need中，同时window[d] == need[d],则valid–，window[d]–）;

复杂度

时间复杂度:

$O(n)$;其中$n$为字符串$s$的长度

空间复杂度:

$O(n)$

Code

class Solution {
public:
    /**
     * Two pointer
     *
     * @param s Given string
     * @param t Given string
     * @return string
     */
    string minWindow(string s, string t) {
        unordered_map<char, int> need;
        unordered_map<char, int> window;
        for (char c: t) {
            need[c]++;
        }
        int left = 0;
        int right = 0;
        int valid = 0;
        // Records the starting index and length of the minimum overlay substring
        int start = 0;
        int len = INT_MAX;
        while (right < s.size()) {
            //c is the character moved into the window
            char c = s[right];
            // Move the window right
            right++;
            // Perform some column updates to the data in the window
            if (need.count(c)) {
                window[c]++;
                if (window[c] == need[c]) {
                    valid++;
                }
            }
            // Determine whether to shrink the left window
            while (valid == need.size()) {
                // Update the minimum overlay substring
                if (right - left < len) {
                    start = left;
                    len = right - left;
                }
                //d is the character to be moved out of the window
                char d = s[left];
                // Move the window left
                left++;
                // Perform some column updates to the data in the window
                if (need.count(d)) {
                    if (window[d] == need[d]) {
                        valid--;
                    }
                    window[d]--;
                }
            }
        }
        // Returns the minimum overlay substring
        return len == INT_MAX ? "" : s.substr(start, len);
    }
};