题目描述
罗马数字包含以下七种字符: I, V, X, L,C,D 和 M。
示例
输入:“III”
输出:3
详情搜索leetcode题目
代码
public int romanToInt(String s) {
int answer = 0;
for (int i = 0; i < s.length(); i++) {
if (s.charAt(i) == 'M')
answer += 1000;
if (s.charAt(i) == 'C') {
if (i + 1 < s.length() && s.charAt(i + 1) == 'M') {
answer -= 100;
} else if (i + 1 < s.length() && s.charAt(i + 1) == 'D') {
answer -= 100;
} else
answer += 100;
}
if (s.charAt(i) == 'D') {
answer += 500;
}
if (s.charAt(i) == 'X') {
if (i + 1 < s.length() && s.charAt(i + 1) == 'C') {
answer -= 10;
} else if (i + 1 < s.length() && s.charAt(i + 1) == 'L') {
answer -= 10;
} else
answer += 10;
}
if (s.charAt(i) == 'L')
answer += 50;
if (s.charAt(i) == 'I') {
if (i + 1 < s.length() && s.charAt(i + 1) == 'X') {
answer -= 1;
} else if (i + 1 < s.length() && s.charAt(i + 1) == 'V') {
answer -= 1;
} else
answer += 1;
}
if (s.charAt(i) == 'V')
answer += 5;
}
return answer;
}
第一次想的代码还没进行优化,太多的if else判断 程序效率较低,第二次我们换成了(字符串的底层就是数组)数组 和 switch 关键是switch提高了程序的效率
程序如下
public int romanToInt(String s) {
if("".equals(s))
return 0;
char [] strlen = s.toCharArray();
int answer = 0;
for (int i = 0; i < strlen.length; i++) {
switch (strlen[i]) {
case 'M':
answer += 1000;
break;
case 'C':
if (i + 1 < strlen.length && strlen[i+1] == 'M') {
answer -= 100;
} else if (i + 1 < strlen.length && strlen[i+1] == 'D') {
answer -= 100;
} else {
answer += 100;
}
break;
case 'D':
answer += 500;
break;
case 'X':
if (i + 1 < strlen.length && strlen[i+1] == 'C') {
answer -= 10;
} else if (i + 1 < strlen.length && strlen[i+1] == 'L') {
answer -= 10;
} else {
answer += 10;
}
break;
case 'L':
answer += 50;
break;
case 'I':
if (i + 1 < strlen.length && strlen[i+1] == 'X') {
answer -= 1;
} else if (i + 1 < strlen.length &&strlen[i+1] == 'V') {
answer -= 1;
} else {
answer += 1;
}
break;
case 'V':
answer += 5;
break;
}
}
return answer;
}
代码判断没有改变,换成switch后 提升50ms达到了95%可见高效