【leetcode】67. Add Binary

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標籤:leetcode   addbinary   algorithm   

@requires_authorization@author johnsondu@create_time 2015.7.15 11:00@url [add binary](https://leetcode.com/problems/add-binary/)/******************* *  類比大數相加 *  時間複雜度: O(n) *  空間複雜度: O(n) ******************/class Solution {public:    string addBinary(string a, string b) {        int lena = a.size();        int lenb = b.size();        string ans = "";        string tmpa = "";        string tmpb = "";        for(int i = lena-1; i >= 0; i --) tmpa += a[i];        for(int i = lenb-1; i >= 0; i --) tmpb += b[i];        int mins = min(lena, lenb);        int carry = 0;        for(int i = 0; i < mins; i ++){            int res = (tmpa[i] - ‘0‘) + (tmpb[i] - ‘0‘) + carry;             carry = res > 1 ? 1 : 0;            res = res % 2;            ans += (res + ‘0‘);        }        for(int i = mins; i < lena; i ++){            int res = (tmpa[i] - ‘0‘) + carry;            carry = res > 1 ? 1 : 0;            res = res % 2;            ans += (res + ‘0‘);        }        for(int i = mins; i < lenb; i ++){            int res = (tmpb[i] - ‘0‘) + carry;            carry = res > 1 ? 1 : 0;            res = res % 2;            ans += (res + ‘0‘);        }        if(carry){            ans += (carry + ‘0‘);        }        int len_ans = ans.size();        for(int i = 0; i < len_ans / 2; i ++){            char tmp = ans[i];            ans[i] = ans[len_ans-i-1];            ans[len_ans-i-1] = tmp;        }        return ans;    }};
// Simplified Versionclass Solution {public:    string addBinary(string a, string b) {        int lena = a.size() - 1;        int lenb = b.size() - 1;        string ans = "";        int carry = 0;        while(lena >= 0 || lenb >= 0 || carry > 0){            int res = carry;            if(lena >= 0) res += (a[lena] - ‘0‘);            if(lenb >= 0) res += (b[lenb] - ‘0‘);            carry = res / 2;            ans = string(1, (res & 1) + ‘0‘) + ans;            lena --;            lenb --;        }        return ans;    }};

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【leetcode】67. Add Binary

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