[LeetCode-interview algorithm classic-Java implementation] [067-Add Binary (Binary addition)], leetcode -- java

[LeetCode-interview algorithm classic-Java implementation] [067-Add Binary (Binary addition)], leetcode -- java
[067-Add Binary (Binary addition )][LeetCode-interview algorithm classic-Java implementation] [directory indexes for all questions]Original question

Given two binary strings, return their sum (also a binary string ).
For example,
A ="11"
B ="1"
Return"100"

Theme

Given two binary strings, return their sum, which is also a binary string.

Solutions

First, convert the two binary strings to the corresponding integer array, and add them from the low position to the high position. At the same time, consider the situation that the last addition also needs to expand one bit. For more information, see code implementation.

Code Implementation

Algorithm Implementation class

Public class Solution {public String addBinary (String a, String B) {int [] ca = new int [. length ()]; int [] cb = new int [B. length ()]; // converts the value in the character array to 0 or 1 for (int I = 0; I <. length (); I ++) {ca [I] =. charAt (I)-'0';} // converts the value in the character array to 0 or 1 for (int I = 0; I <B. length (); I ++) {cb [I] = B. charAt (I)-'0';} // if (ca. length <cb. length) {int [] tmp = ca; ca = cb; cb = tmp;} int ai = ca. length-1; // character Array ca's last index subscript int bi = cb. length-1; // The character array cb's last index subscript int carry = 0; // The lower-bit carry mark int result; // The loaded result // calculation example: 1010101101 + 10100 while (ai> = 0 & bi> = 0) {result = ca [ai] + cb [bi] + carry; ca [ai] = result % 2; carry = result/2; ai --; bi --;} // process the remaining number while (ai> = 0) {result = ca [ai] + carry; ca [ai] = result % 2; carry = result/2; if (carry = 0) {break;} ai --;} // convert the value in the character array to 0 or 1 for (int I = 0; I <ca. length; I ++) {ca [I] + = '0';} // you do not need to extend an if (carry = 0) {char [] ch = new char [ca. length]; for (int I = 0; I <ca. length; I ++) {ch [I] = (char) (ca [I]);} return new String (ch );} // an else {char [] ch = new char [ca. length + 1]; ch [0] = '1'; for (int I = 0; I <ca. length; I ++) {ch [I + 1] = (char) (ca [I]);} return new String (ch );}}}

Evaluation Result

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