[LeetCode]13. Roman to Integer

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Given a roman numeral, convert it to an integer.

Input is guaranteed to be within the range from 1 to 3999.

思路：建立一个与字符串等长的数组存放标志位，遍历整个字符串，若该的字符所表示的数字比其右边相邻的字符表示的数字小，则标志位记为-1；

否则记为1

将对应的字符表示的数字与相应的标志位相乘，做和得出结果

     <span style="white-space:pre"></span>public int romanToInt(String s) {<span style="white-space:pre"></span>if (s == null) return 0;<span style="white-space:pre"></span><span style="white-space:pre"></span>char [] temp = s.toCharArray();<span style="white-space:pre"></span>int [] it = new int[temp.length];<span style="white-space:pre"></span>int [] a = new int[temp.length];<span style="white-space:pre"></span>int sum = 0;<span style="white-space:pre"></span>final  char I = 'I';<span style="white-space:pre"></span>final  char X = 'X';<span style="white-space:pre"></span>final  char L = 'L';<span style="white-space:pre"></span>final  char C = 'C';<span style="white-space:pre"></span>final  char D = 'D';<span style="white-space:pre"></span>final  char M = 'M';<span style="white-space:pre"></span>final  char V = 'V';<span style="white-space:pre"></span>for (int i = 0; i < temp.length; i++) {<span style="white-space:pre"></span>char t = temp[i];<span style="white-space:pre"></span>switch(t) {<span style="white-space:pre"></span>case I : it[i] = 1; break;<span style="white-space:pre"></span>case V : it[i] = 5; break;<span style="white-space:pre"></span>case X : it[i] = 10;break;<span style="white-space:pre"></span>case L : it[i] = 50; break;<span style="white-space:pre"></span>case C : it[i] = 100; break;<span style="white-space:pre"></span>case D : it[i] = 500; break;<span style="white-space:pre"></span>case M : it[i] = 1000; break;<span style="white-space:pre"></span>}<span style="white-space:pre"></span>for (int j = 0; j < it.length-1; j++) {<span style="white-space:pre"></span>if (it[j] < it[j+1]) a[j] = -1;<span style="white-space:pre"></span>else a[j] = 1;<span style="white-space:pre"></span><span style="white-space:pre"></span>}<span style="white-space:pre"></span>}<span style="white-space:pre"></span><span style="white-space:pre"></span>a[temp.length-1] = 1;<span style="white-space:pre"></span>for(int m = 0;m < temp.length; m++) {<span style="white-space:pre"></span>sum = sum + it[m]*a[m];<span style="white-space:pre"></span>}<span style="white-space:pre"></span>return sum;<span style="white-space:pre"></span><span style="white-space:pre"></span>}

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