public class Solution {
public int romanToInt(String s) {
if(s.length() == 0) return 0;
int num = getNum(s.charAt(0));
int prev = num;
for(int i = 1; i < s.length(); i++){
int cur = getNum(s.charAt(i));
if(cur > prev){
num -= 2 * prev;
}
prev = cur;
num += cur;
}
return num;
}
private int getNum(char chr){
switch(chr){
case 'I':
return 1;
case 'V':
return 5;
case 'X':
return 10;
case 'L':
return 50;
case 'C':
return 100;
case 'D':
return 500;
case 'M':
return 1000;
default:
return 0;
}
}
}
这题怪怪的,我也不知道到底想考啥。。难道是迭代和递归么。。。。
public class Solution {
public String countAndSay(int n) {
if(n == 1) return "1";
StringBuilder sb = new StringBuilder();
sb.append("1");
for(int i = 1; i < n; i++){
String str = sb.toString();
sb.setLength(0);
int index = 0;
while(index < str.length()){
int num = str.charAt(index) - '0';
int count = 1;
while(index < str.length() - 1 && str.charAt(index) == str.charAt(index + 1)){
count ++;
index ++;
}
index ++;
sb.append(count);
sb.append(num);
}
}
return sb.toString();
}
}
之前那道 Roman to Integer 就弄了个 switch case ,这次可能性稍微多了点,直接开两个 1-1 onto mapping 当表查好了。