LeetCode 43.字符串相乘 模拟

题意：给两个十进制字符串，求出两数相乘得到的结果字符串
思路：字符串长度太大，而且用BigInteger做这题就没意思了，所以只能按照乘法法则模拟。首先模拟两字符串数相加过程，之后模拟多位和个位相乘过程，之后按照位数往后添0相加得到总和就是结果

class Solution {
    public String multiply(String num1, String num2) {
        if (num1.equals("0") || num2.equals("0"))
            return "0";
        int j = num2.length()-1;
        String ans = "0", s2;
        int cnt = 0;
        while (j >= 0) {
        	s2 = "";
        	int y = num2.charAt(j)-'0', flag = 0;
        	for (int i = num1.length()-1; i >= 0; i--) {
        		int x = num1.charAt(i)-'0', num = x*y;
        		if (num%10+flag >= 10) {
        			s2 = (num%10+flag-10) + s2;
        			flag = num/10+1;
        		} else {
        			s2 = (num%10+flag) + s2;
        			flag = num/10;
        		}
        	}
        	if (flag != 0)
        		s2 = flag + s2;
        	for (int i = 0; i < cnt; i++)
        		s2 += "0";
        	cnt++;
        	ans = add(ans, s2);
        	j--;
        }
        return ans;
    }
	public String add(String num1, String num2) {
		int i = num1.length()-1, j = num2.length()-1;
		String ans = "";
		boolean flag = false;
		while (i >= 0 || j >= 0) {
			if (i < 0) {
				int x = num2.charAt(j)-'0';
				if (flag && x+1 >= 10) {
					ans = (x-9) + ans;
				} else if (flag) {
					ans = (x+1) + ans;
					flag = false;
				} else {
					ans = x + ans;
				}
			} else if (j < 0) {
				int x = num1.charAt(i)-'0';
				if (flag && x+1 >= 10) {
					ans = (x-9) + ans;
				} else if (flag) {
					ans = (x+1) + ans;
					flag = false;
				} else {
					ans = x + ans;
				}
			} else {
				int x = num1.charAt(i)-'0', y = num2.charAt(j)-'0', sum = x + y;
				if (sum >= 10) {
					if (flag) {
						ans = (sum-9) + ans;
					} else {
						ans = (sum-10) + ans;
						flag = true;
					}
				} else {
					if (flag && sum+1 >= 10) {
						ans = (sum-9) + ans;
					} else if (flag) {
						ans = (sum+1) + ans;
						flag = false;
					} else {
						ans = sum + ans;
					}
				}
			}
			i--; j--;
		}
		if (flag) {
			ans = "1" + ans;
		}
		return ans;
	}
}