从零开始的LC刷题(10): Implement strStr() 字符串匹配

原题:

Implement strStr().

Return the index of the first occurrence of needle in haystack, or -1 if needle is not part of haystack.

Example 1:

Input: haystack = "hello", needle = "ll"
Output: 2

Example 2:

Input: haystack = "aaaaa", needle = "bba"
Output: -1

字符串匹配问题,四种算法,先从暴力匹配算法开始:

1)暴力匹配:速度极慢,在匹配失败时会从下一位置继续匹配,浪费大量时间O(mn)

结果:

Success

Runtime: 1292 ms, faster than 15.84% of C++ online submissions for Implement strStr().

Memory Usage: 8.8 MB, less than 100.00% of C++ online submissions for Implement strStr().

代码:

class Solution {
public:
    int strStr(string haystack, string needle) {
        int size=haystack.size();
        int size2=needle.size();
        if(size2==0){return 0;}
        if(size==0){return -1;}
        for(int i=0;i<size;i++){
            int x=0,y=i;
            while(haystack[y]==needle[x]&&x<size2){
                x++;
                y++;
            }
            if(x==size2){return i;}
            
        }
        return -1;
    }
};

(2)改进暴力匹配:使用哈希表,在哈希值相同的时候再暴力匹配,速度有一定提升。本来一开始打算写个hash函数,结果性能提升只有200ms不说内存占用还达到了600多mb,吓到我了,肯定是因为参数传递导致的。所以我又把hash函数内置了,然后性能有了极大提升,结果如下:

Success

Runtime: 4 ms, faster than 100.00% of C++ online submissions for Implement strStr().

Memory Usage: 8.9 MB, less than 99.61% of C++ online submissions for Implement strStr().

代码:

class Solution {
public:
    int strStr(string haystack, string needle) {
        int size=haystack.size();
        int size2=needle.size();
        int ret = 0;
        for (int i=0;i<size2;i++){ret += needle[i];}
        if(size2==0){return 0;}
        if(size==0){return -1;}
        for(int i=0;i<size;i++){
            if(size-i<size2){return -1;}
            int x=0,y=i;
            int ret2 = 0;
            for (int j=i;j<size2+i;j++){ret2 += haystack[j];}
            if(ret==ret2){
                while(haystack[y]==needle[x]&&x<size2){
                    x++;
                    y++;
                    
                }
                if(x==size2){return i;}
            }
            
            
        }
        return -1;
    }
};

这就4ms了但是更好的算法还没用呢让我情何以堪啊,好像时间单位就是4ms这尼玛没有提升空间了啊。

(3)KMP:经典匹配算法,原理可以百度,最关键的是求next数组

结果:

Success

Runtime: 4 ms, faster than 100.00% of C++ online submissions for Implement strStr().

Memory Usage: 9.2 MB, less than 98.25% of C++ online submissions for Implement strStr().

代码:

class Solution {
public:
    int strStr(string haystack, string needle) {
        int size1=haystack.size();
        int size2=needle.size();
        if(size2==0){return 0;}
        if(size1==0){return -1;}
        int* next=(int*)malloc(size2*sizeof(int));
	    next[0] = -1;
	    int k = -1;
	    int l = 0;
	    while (l < size2 - 1)
	    {
		    if (k == -1 || needle[l] == needle[k]) 
		    {
			    ++k;
			    ++l;
			    next[l] = k;
		    }
		    else {
			k = next[k];
		    }
	    }

        int i=0,j=0;
        while(i<size1&&j<size2){
            if(j==-1||haystack[i]==needle[j]){
                i++;
                j++;
            }
            else{j=next[j];}
        }
        if(j==needle.size()){return i-j;}
        else return -1;
    }
};

(4)BM算法:这个算法是真的麻烦,以后再写把哈哈哈哈哈???

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转载自blog.csdn.net/cyr429/article/details/89491145