Problem: 438. 找到字符串中所有字母异位词

题目描述

思路及解法

1.编写辅助函数bool same(vector& need, vector& matched)：

1.1 以need为标准，循环对比need和matched的每一个位置的元素值是否相等

2.获取s和p的长度，lenS和lenP;
3.创建vector needs(26,0);，将p中的字符映射到其中needs[p[i] - ‘a’]++;
4.创建窗口：创建vector matched(26,0);,并定义指针starP、endP，并且将字符串s中的前lenP个字符映射到matched中,并调用same进行一次判断，若满足则将startP添加到结果集合result中
5.维护窗口：当两个指针均小于lenS时执行**matched[s[starP] - ‘a’]–;matched[s[endP] - ‘a’]++;**并且让startP与endP自增，并调用same函数，将字母异位置词的起始位置添加到result中

复杂度

时间复杂度:

$O(1)$

空间复杂度:

$O(1)$

Code

class Solution {
public:
    /**
     * Sliding window
     *
     * @param s The string given to be matched
     * @param p Substring
     * @return vector<int>
     */
    vector<int> findAnagrams(string s, string p) {
        int sLen = s.length();
        int pLen = p.length();
        if (pLen > sLen) {
            return {};
        }
        vector<int> needs(26,0);
        for (int i = 0; i < pLen; ++i) {
            needs[p[i] - 'a']++;
        }
        vector<int> matched(26,0);
        int starP = 0;
        int endP = 0;
        vector<int> result;
        while (endP < pLen) {
            matched[s[endP] - 'a']++;
            endP++;
        }
        if (same(needs, matched)) {
            result.push_back(starP);
        }
        while (starP < sLen && endP < sLen) {
            matched[s[starP] - 'a']--;
            matched[s[endP] - 'a']++;
            starP++;
            endP++;
            if (same(needs, matched)) {
                result.push_back(starP);
            }
        }
        return result;
    }
private:
    /**
     *Determines whether the values in two arrays (within a given range) are equal
     * 
     * @param need Judgment reference array
     * @param matched Array to match each time
     * @return bool
     */
    bool same(vector<int>& need, vector<int>& matched) {
        for (int i = 0; i < need.size(); ++i) {
            if (need[i] != matched[i]) {
                return false;
            }
        }
        return true;
    }
};