数据结构 : 字符串

字符串匹配

1. brute force

  char *strstr(const char *str1, const char *str2)
{
    char *cp = (char*)str1;
    char *s1, *s2;
 
    if (*str2 == '\0')
        return ((char *)str1);
 
    while (*cp)
    {
        s1 = cp;
        s2 = (char *) str2;
 
        while (*s1 && *s2 && !(*s1 - *s2))
            s1++, s2++;
 
        if (!*s2)
            return(cp);
 
        cp++;
    }
    return(NULL);
}

2. KMP

void getnext(const char* str, int* next) {
        int k = -1 , i = 0;
        next[0] = -1;
        int len = strlen(str); 
        while ( i < len) {
            if (k == -1 || str[i] == str[k] ) {
                next[++i] = ++k;
            }
            else {
                k = next[k];
            }
        }
    }

    int strStr(const char* source, const char* target, int* next) {
        if (source == NULL || target == NULL)
            return -1;
        if (!(*source) && !(*target))
            return 0;
        getnext(target, next);
        int i = 0, k = 0;
        int slen = strlen(source);
        int tlen = strlen(target); 
        while ( i < slen && k < tlen) {
            if (k == -1 || source[i] == target[k]) {
                ++i; ++k;
            }
            else {
                k = next[k];
            }
        }
        return target[k] ? -1 : i - k;
    }

字符串替换(变长)

题意 ：请实现一个函数，将一个字符串中的空格替换成“%20”。例如，当字符串为We Are Happy.
则经过替换之后的字符串为We%20Are%20Happy。

class Solution {
public:
    void replaceSpace(char *str,int length) {
        if (length == 0){
            return;
        }
           int num = 0;
        for (int i = 0; i < length; ++i){
            if (str[i] == ' '){
                num++;
            }
        }
        int newLength = length + 2 * num;
        for (int i = length - 1, j = newLength - 1; i >=0 || j >= 0;){
            if (str[i] == ' '){
                str[j--] = '0';
                str[j--] = '2';
                str[j--] = '%';
                i--;
            } else{
                str[j--] = str[i--];
            }
        }
        str[newLength] = '\0'; 
    }
};

字符串删除（变短）

题意：删除word和word之间的多余空格，行首和行尾不可以有多余空格

static void delExtraSpace(char *s){
    int len = strlen(s); 
    int newLen = 0;
    for(int i = 0, j = 0; i < len;){
        while (i < len && s[i] == ' ' ){ // 多余空格删除
            i++;
        }
        while (i < len && s[i] != ' '){
            s[j++] = s[i++];
            newLen = j; 
        }
        if (i < len){ 
            s[j++] = s[i++]; // 空格
        }
    }
    s[newLen] = '\0'; 
}

字符串逆置

Given an input string, reverse the string word by word.
For example,
Given s = "the sky is blue",
return "blue is sky the".

static void reverse(int low, int high, char* s){
    for(;low < high; low++, high--){
        s[low] ^= s[high];
        s[high] ^= s[low];
        s[low] ^= s[high]; 
    }
}
void reverseWords(char *s) {
    int len = strlen(s); 
    reverse(0, len - 1, s); 
    for(int low = -1, i = 0; i < len; ++i){
        if (s[i] == ' '){
            if (low != -1){
                reverse(low, i - 1, s); 
                low = -1;
            }
        } else if (low == -1){
            low = i; 
        }
        if (i == len - 1 && low != -1){ // easy to lost this part
            reverse(low, i, s); 
        }
    }
    delExtraSpace(s); 
}

字符串 四则运算

double proccess(string& s) {
    stringstream sstr("+" + s + "+");
    double result = 0;
    double partial = 0;
    double num;
    char op;
    while (sstr >> op) {
        switch (op) {
            case '+':
            case '-':
                result += partial;
                sstr >> partial; 
                partial *= (op == '+' ? 1 : -1);    
            break;
            case '*':
                sstr >> num;
                partial *= num;
            break;
            case '/':
                sstr >> num;
                partial /= num;
        }
    }
    return result;
}

atoi

int myAtoi(char* str) {
    if (str == NULL) {
        return 0;
    }
    while (*str != '\0' && *str == ' '){
        ++str;
    }
    int signal = 1;
    if (*str == '-') {
        signal = -1;
        ++str;
    } else if (*str == '+') {
        ++str;
    }
    int res = 0;
    while (*str != '\0') {
        if (*str >= '0' && *str <= '9') {
            res = res * 10;
            res += signal * (*str - '0');
        } else {
            break;
        }
        ++str;
    }
    return res; 
}