LeetCode 8. String to Integer (atoi)

最新推荐文章于 2020-04-04 20:44:24 发布

原创最新推荐文章于 2020-04-04 20:44:24 发布 · 134 阅读

0 ·

CC 4.0 BY-SA版权

LeetCode 题解专栏收录该内容

288 篇文章

订阅专栏

本文介绍了一个名为myAtoi的函数实现，该函数将输入的字符串转换为整数。需要注意的是，此函数需要处理空白字符、正负号，并且在转换过程中要避免整数溢出的问题。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

There are quite a lot of details to pay attentions to. But, the most important one is : watch out overflow..... Thus, the res value should be long long int type.

Better to compare with this question : http://blog.youkuaiyun.com/github_34333284/article/details/51171908

    int myAtoi(string str) {
            if(str.size() == 0) return 0;
    int i = 0;
    long long int res = 0;
    bool negative = false;
    while(i < str.size() && str[i] == ' ') {
        i++;
    }
    if(str[i] == '-' || str[i] == '+') {str[i++] == '-' ? negative = true : negative = false;}
    while(i < str.size()) {
        if(str[i] >= '0' && str[i] <= '9') {
            res = res * 10 + (str[i] - '0');
            if(res > INT_MAX) return negative ? INT_MIN : INT_MAX;
        } else {break;}
        i++;
    }
    return negative ? -1 * res : res;
    }