本文介绍了一种使用O(n)时间复杂度和恒量空间复杂度来比较两个字符串的方法。通过逐字符对比的方式，直到找到不匹配的字符或者到达字符串的结尾。
比较两个字符串，用O(n)时间和恒量空间。
ANSWER:
What is “comparing two strings”? Just normal string comparison? The natural way use O(n) time and O(1) space.
int strcmp(char * p1, char * p2) {
while (*p1 != ‘\0’ && *p2 != ‘\0’ && *p1 == *p2) {
p1++, p2++;
}
if (*p1 == ‘\0’ && *p2 == ‘\0’) return 0;
if (*p1 == ‘\0’) return -1;
if (*p2 == ‘\0’) return 1;
return (*p1 - *p2); // it can be negotiated whether the above 3 if’s are necessary, I don’t like to omit them.
}
ANSWER:
What is “comparing two strings”? Just normal string comparison? The natural way use O(n) time and O(1) space.
int strcmp(char * p1, char * p2) {
while (*p1 != ‘\0’ && *p2 != ‘\0’ && *p1 == *p2) {
p1++, p2++;
}
if (*p1 == ‘\0’ && *p2 == ‘\0’) return 0;
if (*p1 == ‘\0’) return -1;
if (*p2 == ‘\0’) return 1;
return (*p1 - *p2); // it can be negotiated whether the above 3 if’s are necessary, I don’t like to omit them.
}
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