> 字符串查找匹配题目，数据结构字符串章节本科时候讲过该题目 - 第一种方法是循环匹配，如果不相同就调一位继续全部匹配。效率非常低 - 第二种方法是，当发现不匹配后，不再从头开始继续匹配，而是从原有字符串中找到首字符相同的进行匹配。 - 第三种是参考答案，思路基本相同，但是精简了计算，效率猛增

MyCode1.cpp

//Runtime: 1632 ms, faster than 5.88%
//Memory Usage: 9.6 MB, less than 26.21%

class Solution {
public:
    int strStr(string haystack, string needle) {
        if(needle.length() == 0) return 0;
        
        int res;
        
        for(int i = 0;i < haystack.length();++i){
            int j = 0;
            res = i;
            while(haystack[i] == needle[j]){
                // cout << "i = " << i << " " << "j = " << j << endl;
                j++;i++;
                if(j == needle.length())
                    return i - needle.length();
            }
            i = res;
        }
        return -1;
    }
};

MyCode2.cpp

//Runtime: 2708 ms, faster than 5.01% 
//Memory Usage: 9.6 MB, less than 38.45%

class Solution {
public:
    int strStr(string haystack, string needle) {
        if(needle.length() == 0) return 0;
        
        int res;
        
        for(int i = 0;i < haystack.length();++i){
            int j = 0;
            res = i;
            bool m = true;
            while(haystack[i] == needle[j]){
                cout << "i = " << i << " " << "j = " << j << endl;
                j++;i++;
                if(haystack[res] == haystack[i] && m == true){
                    res = i - 1;
                    m = false;
                }
                    
                if(j == needle.length())
                    return i - needle.length();
            }
            i = res;
        }
        return -1;
    }
};

refer1.cpp

//Runtime: 8 ms, faster than 99.35%
//Memory Usage: 9.5 MB, less than 42.72%

class Solution {
public:
    int strStr(string haystack, string needle) {
        int m = haystack.size(),n = needle.size();
        for(int i = 0;i <= m -n;i++){
            int j = 0;
            for(;j < n;j++){
                if(haystack[i + j] != needle[j])
                    break;
            }
            if(j == n)
                return i;
        }
        return -1;
    }
};