我接受了一次非常糟糕的采访。所以,现在我试图找到问题的解决方案。这是面试问题:
"我们有以下映射:
M:1000,D:500,C:100,L:50,X:10,V:5,I:1。
我们有以下规则:
每个字母都映射为正整数值
您可以将值添加到一起,但是...
...当一个值(或相同值的运行)后面跟一个更大的值时,你减去该一组值的总和。
示例:
IIX - > 8
MCCMIIX - > 1808
我们获得了这种Java方法:int valueOfRoman(char roman)
。
我们已经实现了Java方法:int romanToInt(String s)
"
我知道这不是一个合适的罗马数字系统,但这是实际的问题。
我能够为正确的罗马系统编写一个有效的解决方案。但是我无法改变它以便它适应这些新规则,特别是规则3.我已经尝试过,但没有成功。我的解决方案现在的方式是,对于IIX,它打印10,而不是8的正确答案。这是我的代码(我还为我的测试实现了valueOf
):
static int romanToInt(String s) {
char curr;
int currVal;
char prev;
int prevVal;
int total = valueOfRoman(s.charAt(0));
for (int i = 1; i < s.length(); i++) {
curr = s.charAt(i);
currVal = valueOfRoman(curr);
prev = s.charAt(i-1);
prevVal = valueOfRoman(prev);
total += currVal;
if(currVal > prevVal) {
total = total - (2*prevVal);
}
}
return total;
}
static int valueOfRoman(char c) {
if (c == 'M') {
return 1000;
} else if (c == 'D') {
return 500;
} else if (c == 'C') {
return 100;
} else if (c == 'L') {
return 50;
} else if (c == 'X') {
return 10;
} else if (c == 'V') {
return 5;
} else if (c == 'I') {
return 1;
}
return -1;
}
非常感谢任何帮助。特别有用的是,如果你能告诉我如何修改我的代码。谢谢!
编辑:我编辑了方法的名称,以便更清楚。答案 0 :(得分:2)
我接受了您提供的小型测试。
static int rom2int(String s) {
if (s == null || s.length() == 0) {
return 0;
}
// Total value.
int total = 0;
// The most recent.
char current = s.charAt(0);
// Total for the current run.
int run = valueOf(current);
for (int i = 1; i < s.length(); i++) {
char next = s.charAt(i);
int value = valueOf(next);
if (next == current) {
// We're in a run - just keep track of its value.
run += value;
} else {
// Up or down?
if (value < valueOf(current)) {
// Gone down! Add.
total += run;
} else {
// Gone UP! Subtract.
total -= run;
}
// Run ended.
run = valueOf(next);
}
// Kee track of most recent.
current = next;
}
return total + run;
}
private void test(String s) {
System.out.println("Value of " + s + " = " + rom2int(s));
}
public void test() {
test("IVX");
test("IIVVL");
test("IIX");
test("MCCMIIX");
test("MVVV");
}
打印
Value of IVX = 4 - Odd!!!
Value of IIVVL = 38
Value of IIX = 8
Value of MCCMIIX = 1808
Value of MVVV = 1015
答案 1 :(得分:1)
以下是我如何解决问题的方法:
答案 2 :(得分:1)
所以,没有人抓住我的暗示。那我也试一试。我不会进入“IVX” - 因为我认为这是一个语法错误。
int romanToInt( String s ){
int total = 0;
int pivot = 0;
for( int i = s.length()-1; i >= 0; i--){ // We start at the **end**
int current = valueOfRoman((s.charAt(i));
if( current >= pivot ){ // We will have at least "I", so it **will** be > pivot for 1st char.
pivot = current;
total += pivot;
}else{
total -= current;
}
}
return total;
}
让我们看看:IIX
i char value total pivot -> total pivot 2 X 10 0 0 > 10 10 1 I 1 10 10 < 9 10 0 I 1 9 10 < 8 10
MCCMIIX
i char value total pivot -> total pivot 6 X 10 0 0 > 10 10 5 I 1 10 10 < 9 10 4 I 1 9 10 < 8 10 3 M 1000 8 10 > 1008 1000 2 C 100 1008 1000 < 908 1000 1 C 100 908 1000 < 808 1000 0 M 1000 808 1000 = 1808 1000
该方法省略了输入验证以简化。我假设所有输入都已经过检查,并且根据“规则”仅包含允许的字符。
答案 3 :(得分:0)
我接受它。
编辑更改#2
public class Romans {
private int valueOf(char c) {
if (c == 'M') {
return 1000;
} else if (c == 'D') {
return 500;
} else if (c == 'C') {
return 100;
} else if (c == 'L') {
return 50;
} else if (c == 'X') {
return 10;
} else if (c == 'V') {
return 5;
} else if (c == 'I') {
return 1;
}
return 0;
}
public int rom2int(String s) {
int currVal;
int runValue = 0;
int repetition = 0;
int total = 0;
boolean alreadyAdded = false;
for (int i = 0; i < s.length(); i++) {
currVal = valueOf(s.charAt(i));
if (runValue == 0) {
runValue = currVal;
repetition = 1;
alreadyAdded = false;
} else if (currVal > runValue) {
total = total + (currVal - (runValue * repetition));
repetition = 1;
runValue = currVal;
alreadyAdded = true;
} else if (currVal < runValue) {
if(!alreadyAdded) {
total += (runValue * repetition);
}
repetition = 1;
runValue = currVal;
alreadyAdded = false;
} else {
repetition++;
alreadyAdded = false;
}
}
if (!alreadyAdded) {
total += (runValue * repetition);
}
return total;
}
}
主要是我正在运行:
public static void main(String[] args) {
Romans r = new Romans();
String[] inputs = {"MVVV", "IIX","MCCMIIX", "IVX"};
for(String input : inputs) {
System.out.println("Value of " + input + " is: " + r.rom2int(input));
}
}
输出:
Value of MVVV is: 1015
Value of IIX is: 8
Value of MCCMIIX is: 1808
Value of IVX is: 9
答案 4 :(得分:0)
我就这样做了。
它适用于您提到的那两个值(IIX = 8和MCCMIIX = 1808):
public static int rom2int(String s)
{
int currVal = 0, prevVal = 0, subTotal = 0, total = 0;
for (int i = 0; i < s.length(); i++) {
currVal = valueOf(s.charAt(i));
if (currVal > 0) {
if (prevVal == 0) {
subTotal = currVal;
}
else if (currVal > prevVal) {
total += (currVal - subTotal);
subTotal = 0;
}
else if (currVal < prevVal) {
total += subTotal;
subTotal = currVal;
}
else if (currVal == prevVal) {
subTotal += currVal;
}
prevVal = currVal;
}
}
total += subTotal;
return total;
}
public static int valueOf(char c)
{
if (c == 'M')
return 1000;
if (c == 'D')
return 500;
if (c == 'C')
return 100;
if (c == 'L')
return 50;
if (c == 'X')
return 10;
if (c == 'V')
return 5;
if (c == 'I')
return 1;
return 0;
}
编辑1:
“IVX”值的说明:
“...将值加在一起,除非值(或运行SAME) value)后跟一个更大的值,你减去它的总和 价值观。“
IVX = 1 5 10
if 5 > 1 then 5 - 1 = 4
if 10 > 5 then 10 - 0(*) = 10 (*) we already have used V(5) before, so we discard it.
所以IVX的答案是14!
答案 5 :(得分:-2)
这种问题通常很容易用递归的思维方式来解决。解决方案可能如下所示:
public int rom2int(String s)
{
if(s.length() == 0)
// no string --> 0
return 0;
else if(s.length() == 1)
// One Character --> Value of Character
return valueOf(s.charAt(0));
else if((valueOf(s.charAt(0)) > valueOf(s.charAt(1))) )
// The value is NOT followed by a greater value --> We had the value
return rom2int(s.substring(1, s.length())) + valueOf(s.charAt(0));
else if(valueOf(s.charAt(0)) <= valueOf(s.charAt(1)) )
// The value is followed by a greater (or same) value --> We substract the value
return rom2int(s.substring(1, s.length())) - valueOf(s.charAt(0));
else
// Shouldn't Happen. 0 as neutral element in in a Sum.
return 0;
}
即使禁止使用递归解决方案,在我看来,取消递归此算法比首先尝试使用程序错误更简单=)
编辑: 我的结果:
MCCMIIX的价值是:1808
IIX的价值是:8
IVX的价值是:4
IIVVL的价值是:38