Implement atoi to convert a string to an integer.
Hint: Carefully consider all possible input cases. If you want a challenge, please do not see below and ask yourself what are the possible input cases.
Notes: It is intended for this problem to be specified vaguely (ie, no given input specs). You are responsible to gather all the input requirements up front.
這道題的要求還是很多的,很多的case都需要注意,需要滿足要求,例如開頭的空格、開頭的+、-符號標志、字符滿足數字的范圍,並且需要滿足int型整數的正、負邊界。關於符號位的判定,我用了一個bool變量來做標記,關於int型邊界的判定,我采用的方法是在每次乘10做加法之前,先使用最大邊界值做減法然後除以10判定是否會溢出,在不溢出的情況下再繼續進行運算。代碼如下:
class Solution { public: int atoi(const char *str) { bool minus = false; int result = 0; const char *temp = str; unsigned int MAX = 2147483647; unsigned int MIN = 2147483648; while(*temp == ' ') *temp++; if(*temp == '-') { minus = true; temp++; } else if(*temp == '+') { minus = false; temp++; } if(*temp==NULL || *temp<'0' || *temp>'9') return 0; while(*temp >= '0' && *temp <= '9') { if(!minus && (MAX-(int)(*temp-'0'))/10 >= result) { result = result*10+(int)(*temp - '0'); } else if(minus && (MIN-(int)(*temp-'0'))/10 >= result) { result = result*10+(int)(*temp - '0'); } else { return minus ? INT_MIN : INT_MAX; } temp++; } if(minus) result = 0-result; return result; } };