365天挑战LeetCode1000题——Day 064 检查单词是否为句中其他单词的前缀 单词规律 划分字母区间

1455. 检查单词是否为句中其他单词的前缀

代码实现（自解）

class Solution {
public:
    int isPrefixOfWord(string sentence, string searchWord) {
        int pst = 0;
        int n = sentence.size();
        int ss = searchWord.size();
        int sub = 1;
        while (pst != n) {
            if (sentence[pst] == ' ') {
                pst++;
                continue;
            }
            if (sentence.substr(pst, ss) == searchWord) {
                return sub;
            }
            else {
                while (pst != n && sentence[pst] != ' ') pst++;
                sub++;
            }
        }
        return -1;
    }
};

---

290. 单词规律

代码实现（自解）

class Solution {
public:
    bool wordPattern(string s, string pattern) {
        int ps = 0;
        int pp = 0;
        int left, right;
        map<char, string> mapping;
        set<string> st;
        int n = pattern.size();
        while (pp != n) {
            if (pattern[pp] == ' ') {
                pp++;
                continue;
            }
            left = pp;
            right = left;
            while (right != n && pattern[right] != ' ') {
                right++;
            }
            if (!mapping.count(s[ps])) {
                if (st.count(pattern.substr(left, right - left))) return false;
                st.emplace(pattern.substr(left, right - left));
                mapping[s[ps]] = pattern.substr(left, right - left);
            }
            else if (mapping[s[ps]] != pattern.substr(left, right - left))
                return false;
            pp = right;
            ps++;
        }
        if (ps != s.size()) return false;
        return true;
    }
};

---

763. 划分字母区间

代码实现（自解）

class Solution {
public:
    vector<int> partitionLabels(string s) {
        map<char, int> left;
        map<char, int> right;
        vector<int> ans;
        for (char c : s) {
            right[c]++;
        }
        int i = 1;
        for (char c : s) {
            left[c]++;
            right[c]--;
            if (!right[c]) {
                bool flag = true;
                for (auto pa : left) {
                    if (right[pa.first] != 0) {
                        flag = false;
                    }
                }
                if (flag) {
                    ans.push_back(i);
                    i = 0;
                }
            }
            i++;
        }
        return ans;
    }
};