[LeetCode] 415. Add Strings 解题报告

Given two non-negative integers num1 and num2 represented as string, return the sum of num1 and num2.

Note:

1. The length of both num1 and num2 is < 5100.
2. Both num1 and num2 contains only digits 0-9.
3. Both num1 and num2 does not contain any leading zero.
4. You must not use any built-in BigInteger library or convert the inputs to integer directly.

这一题非常简单，记得前面有个和这个类似的题，大概是100号以前。

我的方法就是从右往左算，如果算完，直接把较长的字符串复制下来即可。另外，使用一个boolean变量控制进位就好。

我用的char型数组，主要考虑效率问题，直接String字符串是效率非常低的。

	private static final char ZERO = '0';
	public String addStrings(String num1, String num2) {
		if (num1.length() == 0 || num2.length() == 0) {
			return num1.length() == 0 ? num1 : num2;
		}
		char[] cArr1 = num1.length() >= num2.length() ? num1.toCharArray() : num2.toCharArray();
		char[] cArr2 = num1.length() < num2.length() ? num1.toCharArray() : num2.toCharArray();
		int nCount = 0;
		int nMinLength = cArr2.length;
		int nMaxLength = cArr1.length;

		char[] cArrResult = new char[nMaxLength + 1];
		boolean bUpDigit = false;
		while (nCount < nMaxLength) {
			char c1 = cArr1[cArr1.length - nCount - 1];
			char c2 = nCount < nMinLength ? cArr2[cArr2.length - nCount - 1] : ZERO;
			int nCurrentDigit = (int) (c1 - ZERO) + (int) (c2 - ZERO);
			nCurrentDigit = bUpDigit ? nCurrentDigit + 1 : nCurrentDigit;
			bUpDigit = false;
			if (nCurrentDigit > 9) {
				bUpDigit = true;
				nCurrentDigit -= 10;
			}
			cArrResult[cArrResult.length - nCount - 1] = (char) (nCurrentDigit + ZERO);
			nCount++;
		}
		String strResult = String.valueOf(cArrResult).substring(1);
		return bUpDigit ? 1 + strResult : strResult;
	}