LeetCode String to Integer (atoi)
28894 ワード
class Solution {
public:
int atoi(const char *str) {
if (str == NULL) return 0;
long val =0, pos = 0;
char ch = '\0';
int stage = 0; // 0-initial, 1-sign-collected, 2-digit-collected, 3-end
bool positive = true;
int pos_threshold = INT_MAX / 10;
int neg_threshold = INT_MIN / 10;
while(stage < 3 && (ch = str[pos]) != '\0') {
switch(stage) {
case 0: // initial stage
if (ch == '-' || ch == '+') {
// sign detected
positive = ch == '+';
stage = 1;
pos++;
} else if (ch >= '0' && ch <= '9') {
// digit detected
stage = 2;
} else if (ch == ' ' || ch == '\t' || ch == '
') {
// leading white space
pos++;
} else {
// other chars, invalid str to convert
stage = 3;
}
;break;
case 1: // sign-collected stage
if (ch >= '0' && ch <= '9') {
// digit after sign
stage = 2;
} else {
// other chars, invalid str to convert
stage = 3;
}
;break;
case 2: // number detected stage
if (ch >= '0' && ch <= '9') {
// digits
int pre_val = val;
if (positive) {
val = val * 10 + (ch - '0');
if (pre_val > pos_threshold || val < pre_val) {
stage = 3;
val = INT_MAX;
}
} else {
val = val * 10 - (ch - '0');
if (pre_val < neg_threshold || val > pre_val) {
stage = 3;
val = INT_MIN;
}
}
pos++;
} else {
// other chars, invalid
stage = 3;
}
;break;
}
}
return val;
}
};広い範囲のデータ型を使用していないので、オーバーフローは考慮する必要があります.小さなステータスマシンです.
第2ラウンド:
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.
Update (2015-02-10):The signature of the
C++ function had been updated. If you still see your function signature accepts a const char * argument, please click the reload button to reset your code definition. spoilers alert... click to show requirements for atoi.
Requirements for atoi:
The function first discards as many whitespace characters as necessary until the first non-whitespace character is found. Then, starting from this character, takes an optional initial plus or minus sign followed by as many numerical digits as possible, and interprets them as a numerical value.
The string can contain additional characters after those that form the integral number, which are ignored and have no effect on the behavior of this function.
If the first sequence of non-whitespace characters in str is not a valid integral number, or if no such sequence exists because either str is empty or it contains only whitespace characters, no conversion is performed.
If no valid conversion could be performed, a zero value is returned. If the correct value is out of the range of representable values, INT_MAX (2147483647) or INT_MIN (-2147483648) is returned.
このような問題を見ると少し煩わしいですが、やはりいつものように各状態のジャンプによって書いてもいいです.コードはほぼ一致しています.
1 class Solution {
2 public:
3 int atoi(string str) {
4 int len = str.size();
5 if (len == 0) {
6 return 0;
7 }
8 int val = 0;
9 int pre_val = 0;
10
11 int pos_threshold = INT_MAX/10;
12 int neg_threshold = INT_MIN/10;
13 bool positive = true;
14
15 int pos = 0;
16 char ch = 0;
17 // 0-inital, 1-sign read, 2-number read, 3-end
18 int stage = 0;
19 while (pos < len && stage < 3) {
20 ch = str[pos];
21 switch(stage) {
22 case 0: // inital stage
23 if (ch == '-' || ch == '+') {
24 // sign read
25 stage = 1;
26 positive = ch == '+';
27 pos++;
28 continue;
29 }
30 if (ch == ' ' || ch == '\t') {
31 // leading white space
32 pos++; //skip them
33 continue;
34 }
35 if (ch >= '0' && ch <= '9') {
36 // number read
37 stage = 2;
38 continue;
39 }
40 //other charaters, invalid case
41 stage = 3;
42 break;
43
44 case 1: // sign read stage
45 if (ch >= '0' || ch <= '9') {
46 // number read
47 stage = 2;
48 continue;
49 }
50 // other character, invalid case(if space between sign and digit is not permited)
51 stage = 3;
52 break;
53
54 case 2: // number read case
55 if (!(ch >= '0' && ch <= '9')) {
56 // character is not digit invalid case
57 stage = 3;
58 continue;
59 }
60 pre_val = val;
61 val = val * 10 + (ch - '0') * (positive ? 1:-1);
62 if (positive) {
63 if (pre_val > pos_threshold || val < pre_val) {
64 // overflow
65 val = INT_MAX;
66 stage = 3;
67 continue;
68 }
69 } else {
70 if (pre_val < neg_threshold || val > pre_val) {
71 val = INT_MIN;
72 stage = 3;
73 continue;
74 }
75 }
76 pos++; // next character
77 break;
78 }
79 }
80 return val;
81 }
82 };私はどんなに退屈でそんなに長く書いているのか、それとも状態を乱用してはいけないのか、簡単に書いてください.
class Solution {
public:
int myAtoi(string str) {
int res = 0;
bool neg = false;
int len = str.size();
int stage = 0;
int pos = 0;
char ch = 0;
// skip white spaces
while (pos < len && str[pos] == ' ') {
pos++;
}
// try read sign
if (pos < len && (str[pos] == '-' || str[pos] == '+')) {
neg = str[pos++] == '-';
}
// read digits
while (pos < len && str[pos] >= '0' && str[pos] <= '9') {
int old = res;
if (neg) {
res = res * 10 - (str[pos] - '0');
if (res > old || old < INT_MIN/10) {
res = INT_MIN;
break;
}
} else {
res = res * 10 + str[pos] - '0';
if (res < old || old > INT_MAX/10) {
res = INT_MAX;
break;
}
}
pos++;
}
return res;
}
};