2024/4/6—力扣—简化路径

代码实现：

// 分割'/'得到名字
char **split(const char *s, int *returnSize) {
    int n = strlen(s);
    char **ans = (char **)malloc(sizeof(char *) * n);
    int l = 0, r = 0, len = 0;
    while (r < n) {
        while (r < n && s[r] == '/') {
            r++;
        }
        l = r;
        while (r < n && s[r] != '/') {
            r++;
        }
        if (l < n) {
            ans[len] = (char*)malloc(sizeof(char) * (r - l + 1)); 
            strncpy(ans[len], s + l, r - l);
            ans[len][r - l] = '\0';
            len++;
        }
    }
    *returnSize = len;
    return ans;
}

char *simplifyPath(char *path){
    int namesSize = 0;
    int n = strlen(path);
    char **names = split(path, &namesSize);
    char **stack = (char**)malloc(sizeof(char*) * namesSize);
    int top = -1;

    for (int i = 0; i < namesSize; ++i) {
        if (!strcmp(names[i], "..")) { // 遇到".."
            if (top > -1) {
                top--;
            }
        } else if (!strcmp(names[i], ".")) { // 遇到"." 跳过
            continue;
        } else {
            if (top == -1 || strcmp(stack[top], names[i])) { // 遇到目录名
               stack[++top] = names[i];
            }
        }
    }
    char *ans = (char*)malloc(sizeof(char) * (n + 1));
    int ind = 0;
    if (top == -1) {
        ans[ind] = '/';
        ind++;
    } else {
        for (int i = 0; i <= top; i++) {
            ans[ind++] = '/';
            strcpy(ans + ind, stack[i]);
            ind += strlen(stack[i]);
        }
    }
    ans[ind] = '\0';
    for (int i = 0; i < namesSize; i++) {
        free(names[i]);
    }
    free(names);
    free(stack);
    return ans;
}