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Topic: Sword refers to Offer67. Convert a string to an integer to
achieve stoi; if the
conversion is not possible, return 0: ① str is empty ② only spaces ③ no numbers ④ The first valid character is not a sign or a number
if the value exceeds int Range, return INT_MIN and INT_MAX, which are determined according to the first valid character bit'+' and'-'. If there is no sign, it will be counted as a positive number; -
Ideas:
1. Simulation traversal: time O(n): it needs to traverse at most once, if it cannot be converted, it will end in the middle, space O(1)
class Solution {
public:
int strToInt(string str) {
if (str.empty()) return 0;
int n = str.size();
int i = 0;
while (i < n && str[i] == ' ') ++i;//跳过开头空格;
if (str[i] == ' ') return 0;//str只有空格
int symbol = 0;//标识正负号
if (str[i] == '+') {
symbol = 1;
if (++i < n && !(str[i] >= '0' && str[i] <= '9')) return 0;//正负号后面没数字
}
else if (str[i] == '-') {
symbol = -1;
if (++i < n && !(str[i] >= '0' && str[i] <= '9')) return 0;//正负号后面没数字
}
else if (str[i] >= '0' && str[i] <= '9') ;
else return 0; //第一个有效字符不是'+','-',数字
long res = 0;//结果
int maxVal = INT_MAX;//边界值
for (; i < n; ++i) {
//初始i一定停在第一个数字上
if (str[i] >= '0' && str[i] <= '9') {
res = res * 10 + (str[i] - '0');
if (res > maxVal) {
//我们跳过了符号，都是按正数算的，最后再考虑正负号
if (symbol == 0 || symbol == 1) return INT_MAX;//没出现+-号就默认按正数算
else return INT_MIN;
}
}
else break;//出现数字以外的其他字符就停止
}
return (symbol == 0 || symbol == 1) ? res : (-1) * res;//没出界的int值，根据符号输出正数/负数
}
};
- Summary:
atoi and stoi: the difference lies in the parameters, stoi parameter is string, atoi is const char*, but both can be converted to int, long, long long, just change to i, l, ll.