LeetCode Add Binary

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原题链接在这里: https://leetcode.com/problems/add-binary/

这道题与Plus One相似。都是维护一个currentDigit(%) 和一个carryDigit(/).

Time O(max(m,n)), Space O(max(m,n)), m = a.length(), n = b.length()

Note:1. 注意是从String的尾部算起，所以注意i 和index的关系；

2. 若是算到最后，carryDigit还是1的话就是有一个最后的进位，要在res的首位加起来。

AC Java:

public class Solution {    public String addBinary(String a, String b) {        if(a==null && b==null && a.length()==0 && b.length()==0){            return "";        }                int len1 = a.length();        int len2 = b.length();        StringBuilder res = new StringBuilder();        int carryDigit = 0;        int currentDigit;        int i = 0;        while(i<len1 && i<len2){            currentDigit = (Character.getNumericValue(a.charAt(len1-1-i)) + Character.getNumericValue(b.charAt(len2-1-i)) + carryDigit)%2;            carryDigit = (Character.getNumericValue(a.charAt(len1-1-i)) + Character.getNumericValue(b.charAt(len2-1-i)) + carryDigit)/2;            if(currentDigit == 0){                res.insert(0,"0");            }else{                res.insert(0,"1");            }            i++;        }                if(len1 > len2){            while(i<len1){                currentDigit = (Character.getNumericValue(a.charAt(len1-1-i)) + carryDigit)%2;                carryDigit = (Character.getNumericValue(a.charAt(len1-1-i)) + carryDigit)/2;                if(currentDigit == 0){                    res.insert(0,"0");                }else{                    res.insert(0,"1");                }                i++;            }        }        if(len1 < len2){            while(i<len2){                currentDigit = (Character.getNumericValue(b.charAt(len2-1-i)) + carryDigit)%2;                carryDigit = (Character.getNumericValue(b.charAt(len2-1-i)) + carryDigit)/2;                if(currentDigit == 0){                    res.insert(0,"0");                }else{                    res.insert(0,"1");                }                i++;            }        }        if(carryDigit != 0){                res.insert(0,"1");        }        return res.toString();    }}

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