C++ 解法, 执行用时: 2ms, 内存消耗: 560KB, 提交时间: 2021-12-19
class Solution {
public:
/**
* 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
*
* 比较版本号
* @param version1 string字符串
* @param version2 string字符串
* @return int整型
*/
int compare(string version1, string version2) {
// write code here
char* version1_p = (char*)version1.data();
int version1_len = version1.length();
int version1_idx = 0;
vector<int> version1_nums;
char* version2_p = (char*)version2.data();
int version2_len = version2.length();
int version2_idx = 0;
vector<int> version2_nums;
uint32_t version1_num_tmp=0;
uint32_t version2_num_tmp=0;
while(version1_idx<=version1_len){
if(version1_num_tmp!=0&&version1_idx==version1_len){
version1_nums.push_back(version1_num_tmp);
}
char c = *(version1_p+version1_idx);
version1_idx++;
if(c=='.'){
version1_nums.push_back(version1_num_tmp);
version1_num_tmp = 0 ;
}
else if (c=='0'){
continue;
}
else{
version1_num_tmp=(version1_num_tmp*10)+(c-'0');
}
}
while(version2_idx<=version2_len){
if(version2_num_tmp!=0&&version2_idx==version2_len){
version2_nums.push_back(version2_num_tmp);
}
char c = *(version2_p+version2_idx);
version2_idx++;
if(c=='.'){
version2_nums.push_back(version2_num_tmp);
version2_num_tmp = 0 ;
}
else if (c=='0'){
continue;
}
else{
version2_num_tmp=(version2_num_tmp*10)+(c-'0');
}
}
int maxnums = version1_nums.size()>version2_nums.size()?version1_nums.size():version2_nums.size();
for(int i=0;i<maxnums;i++){
if(version1_nums.size()<maxnums){
version1_nums.push_back(0);
}
if(version2_nums.size()<maxnums){
version1_nums.push_back(0);
}
}
for(int i=0;i<maxnums;i++){
if(version1_nums[i]>version2_nums[i]){
return 1;
}
else if(version1_nums[i]<version2_nums[i]){
return -1;
}
}
return 0;
}
};
C++ 解法, 执行用时: 3ms, 内存消耗: 332KB, 提交时间: 2022-04-04
class Solution {
public:
/**
* 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
*
* 比较版本号
* @param version1 string字符串
* @param version2 string字符串
* @return int整型
*/
int compare(string version1, string version2) {
// write code here
int len1=version1.length();
int len2=version2.length();
int i=0,j=0;
int result=0;
vector<char> arr1,arr2;
while(i<len1&&j<len2){
while(i<len1&&version1[i]!='.'){
if(version1[i]!='0')
arr1.push_back(version1[i]);
i++;
}
while(j<len2&&version2[j]!='.'){
if(version2[j]!='0')
arr2.push_back(version2[j]);
j++;
}
int l1=arr1.size();
int l2=arr2.size();
if(l1>l2){
result=1;
break;
}else if(l1<l2){
result=-1;
break;
}else{
int k1=0,k2=0;
bool label=false;
while(k1<l1){
if(arr1[k1]>arr2[k2]){
result=1;
label=true;
break;
}else if(arr1[k1]<arr2[k2]){
result=-1;
label=true;
break;
}
k1++;
k2++;
}
if(label)
break;
arr1.clear();
arr2.clear();
i++;
j++;//为了跳过刚刚那个‘.’
}
}
if(result==0){
while(i<len1){
if(version1[i]!='0'&&version1[i]!='.'){
result=1;
break;
}
i++;
}
while(j<len2){
if(version2[j]!='0'&&version2[j]!='.'){
result=-1;
break;
}
j++;
}
}
return result;
}
};
C++ 解法, 执行用时: 3ms, 内存消耗: 392KB, 提交时间: 2022-08-03
class Solution {
public:
/**
* 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
*
* 比较版本号
* @param version1 string字符串
* @param version2 string字符串
* @return int整型
*/
int compare(string version1, string version2) {
// write code here
int v1=0;
int v2=0;
int j=0;
for(int i=0;i<=version1.length();i++){
if(i==version1.length()||version1[i]=='.'){
while(j<version2.length()&&version2[j]!='.'){
v2=v2*10+(version2[j]-'0');
j++;
}
if(v1!=v2){return v1<v2?-1:1;}
if(i==version1.length()&&j==version2.length()){return 0;}
if(i==version1.length()){i--;}
j++;
v1=0;v2=0;
}else{
v1=v1*10+(version1[i]-'0');
}
}
return 0;
}
};
C++ 解法, 执行用时: 3ms, 内存消耗: 392KB, 提交时间: 2022-08-02
class Solution {
public:
/**
* 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
*
* 比较版本号
* @param version1 string字符串
* @param version2 string字符串
* @return int整型
*/
int compare(string version1, string version2) {
// write code here
std::vector<std::string> strarry1;
std::vector<std::string> strarry2;
string q=version1;
while(true){
int pos=q.find('.');
string p = q.substr(0,pos);
q = q.substr(pos+1);
strarry1.push_back(p);
if(pos==-1) break;
}
q=version2;
while(true){
int pos=q.find('.');
string p = q.substr(0,pos);
q = q.substr(pos+1);
strarry2.push_back(p);
if(pos==-1) break;
}
int i=0,j=0;
while(i<strarry1.size() && j<strarry2.size()){
string str1=strarry1[i++];
string str2=strarry2[j++];
int a = atoi(str1.c_str());
int b = atoi(str2.c_str());
if(a>b){return 1;}
else if(a<b){return -1;}
}
while(i<strarry1.size()){
string str1=strarry1[i++];
int a = atoi(str1.c_str());
if(a!=0) return 1;
}
while(j<strarry2.size()){
string str1=strarry2[j++];
int a = atoi(str1.c_str());
if(a!=0) return -1;
}
return 0;
}
};
C 解法, 执行用时: 3ms, 内存消耗: 392KB, 提交时间: 2022-07-07
/**
* 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
*
* 比较版本号
* @param version1 string字符串
* @param version2 string字符串
* @return int整型
*
* C语言声明定义全局变量请加上static，防止重复定义
*/
int* char_to_int(int len, char* version){
int tmp=0,value=0;
int* out=calloc(len+1,sizeof(int));
for(int i=0;i<len;i++){
if(version[i]=='.'){
out[value++] = tmp;
tmp=0;
}
else
tmp = tmp*10 + (version[i]-'0');
}
out[value++] = tmp;
out[value]=-1; //增加一个结束标志
return out;
}
int compare(char* version1, char* version2 ) {
// write code here
int len1=strlen(version1);
int len2=strlen(version2);
int len = len1>len2 ? len1 : len2;
int* out1=char_to_int(len1,version1);
int* out2=char_to_int(len2,version2);
int i=0,flag=0,sum=0;;
for(;i<=len;i++){
if(out1[i]==-1){
if(out2[i]!=-1)
flag=1;
break;
}
else if(out2==-1){
if(out1[i]!=-1)
flag=2;
break;
}
else{
if(out1[i]>out2[i])
return 1;
else if(out1[i]<out2[i])
return -1;
}
}
if(flag==1){
for(int j=i+1;j<len;j++) //不能读到-1
sum+=out2[j];
if(sum>0)
return -1;
}
else if(flag==2){
for(int j=i+1;j<len;j++) //不能读到-1
sum+=out1[j];
if(sum>0)
return 1;
}
return 0;
}