LeetCode-13 Roman to Integer
2016-06-18 12:26
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https://leetcode.com/problems/roman-to-integer/
Given a roman numeral, convert it to an integer.
Input is guaranteed to be within the range from 1 to 3999.
1、罗马数字表示方法:
罗马数字采用七个罗马字母作数字、即Ⅰ(1)、X(10)、C(100)、M(1000)、V(5)、L(50)、D(500)。记数的方法:
相同的数字连写,所表示的数等于这些数字相加得到的数,如 Ⅲ=3;小的数字在大的数字的右边,所表示的数等于这些数字相加得到的数,如 Ⅷ=8、Ⅻ=12;小的数字(限于 Ⅰ、X 和 C)在大的数字的左边,所表示的数等于大数减小数得到的数,如 Ⅳ=4、Ⅸ=9;在一个数的上面画一条横线,表示这个数增值 1,000 倍,如
=5000。
2、map,每个元素与前一个元素比较,会有重复运算(88ms)
class Solution {
public:
int romanToInt(string s) {
typedef pair<char,int> p;
map<char,int> m={
p ('I',1),p ('V',5),p ('X',10),p ('L',50),p ('C',100),p ('D',500),p ('M',1000)
};
int len = s.size(),sum = m[s[0]],pre = sum;
for(int i=1;i<len;i++){
if(m[s[i]]<=pre)
sum+=m[s[i]];
else
sum+=m[s[i]]-2*pre;
pre = m[s[i]];
}
return sum;
}
};
3、map,每个元素与后一个元素比较(68ms)
class Solution {
public:
int romanToInt(string s) {
typedef pair<char,int> p;
map<char,int> m={
p ('I',1),p ('V',5),p ('X',10),p ('L',50),p ('C',100),p ('D',500),p ('M',1000)
};
int len = s.size(),sum = 0;
for(int i=0;i<len-1;i++){
if(m[s[i]]<m[s[i+1]])
sum-=m[s[i]];
else
sum+=m[s[i]];
}
return sum + m[s[len-1]];
}
};
4、hash表,每个元素与后一个比较(36ms)
class Solution {
public:
int romanToInt(string s) {
typedef pair<char,int> p;
int m[256];
m['I'] = 1; m['V'] = 5; m['X'] = 10; m['L'] = 50; m['C'] = 100; m['D'] = 500; m['M'] = 1000;
int len = s.size(),sum = 0;
for(int i=0;i<len-1;i++){
if(m[s[i]]<m[s[i+1]])
sum-=m[s[i]];
else
sum+=m[s[i]];
}
return sum + m[s[len-1]];
}
};
5、hash表,每个元素与后一个比较,减少重复计算(32ms,beat98.48%)
class Solution {
public:
int romanToInt(string s) {
int m[256];
m['I'] = 1; m['V'] = 5; m['X'] = 10; m['L'] = 50; m['C'] = 100; m['D'] = 500; m['M'] = 1000;
int len = s.size(),sum = 0,a ,b=m[s[0]];
for(int i=0;i<len-1;i++){
a = m[s[i]];
b = m[s[i+1]];
if(a < b)
sum-=a;
else
sum+=a;
}
return sum + b;
}
};
Given a roman numeral, convert it to an integer.
Input is guaranteed to be within the range from 1 to 3999.
1、罗马数字表示方法:
罗马数字采用七个罗马字母作数字、即Ⅰ(1)、X(10)、C(100)、M(1000)、V(5)、L(50)、D(500)。记数的方法:
相同的数字连写,所表示的数等于这些数字相加得到的数,如 Ⅲ=3;小的数字在大的数字的右边,所表示的数等于这些数字相加得到的数,如 Ⅷ=8、Ⅻ=12;小的数字(限于 Ⅰ、X 和 C)在大的数字的左边,所表示的数等于大数减小数得到的数,如 Ⅳ=4、Ⅸ=9;在一个数的上面画一条横线,表示这个数增值 1,000 倍,如
=5000。
2、map,每个元素与前一个元素比较,会有重复运算(88ms)
class Solution {
public:
int romanToInt(string s) {
typedef pair<char,int> p;
map<char,int> m={
p ('I',1),p ('V',5),p ('X',10),p ('L',50),p ('C',100),p ('D',500),p ('M',1000)
};
int len = s.size(),sum = m[s[0]],pre = sum;
for(int i=1;i<len;i++){
if(m[s[i]]<=pre)
sum+=m[s[i]];
else
sum+=m[s[i]]-2*pre;
pre = m[s[i]];
}
return sum;
}
};
3、map,每个元素与后一个元素比较(68ms)
class Solution {
public:
int romanToInt(string s) {
typedef pair<char,int> p;
map<char,int> m={
p ('I',1),p ('V',5),p ('X',10),p ('L',50),p ('C',100),p ('D',500),p ('M',1000)
};
int len = s.size(),sum = 0;
for(int i=0;i<len-1;i++){
if(m[s[i]]<m[s[i+1]])
sum-=m[s[i]];
else
sum+=m[s[i]];
}
return sum + m[s[len-1]];
}
};
4、hash表,每个元素与后一个比较(36ms)
class Solution {
public:
int romanToInt(string s) {
typedef pair<char,int> p;
int m[256];
m['I'] = 1; m['V'] = 5; m['X'] = 10; m['L'] = 50; m['C'] = 100; m['D'] = 500; m['M'] = 1000;
int len = s.size(),sum = 0;
for(int i=0;i<len-1;i++){
if(m[s[i]]<m[s[i+1]])
sum-=m[s[i]];
else
sum+=m[s[i]];
}
return sum + m[s[len-1]];
}
};
5、hash表,每个元素与后一个比较,减少重复计算(32ms,beat98.48%)
class Solution {
public:
int romanToInt(string s) {
int m[256];
m['I'] = 1; m['V'] = 5; m['X'] = 10; m['L'] = 50; m['C'] = 100; m['D'] = 500; m['M'] = 1000;
int len = s.size(),sum = 0,a ,b=m[s[0]];
for(int i=0;i<len-1;i++){
a = m[s[i]];
b = m[s[i+1]];
if(a < b)
sum-=a;
else
sum+=a;
}
return sum + b;
}
};
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