3632. Binary sum

Here are two binary strings , Returns the sum of them （ In binary ）. Input is Non empty String and contains only numbers  1  and  0.    Example  1: Input : a = "11", b = "1" Output : "100" Example  2: Input : a = "1010", b = "1011" Output : "10101"    Tips ： Each string consists only of characters  '0'  or  '1'  form . 1 <= a.length, b.length <= 10^4 If the string is not  "0" , No leading zeros . source ： Power button （LeetCode） link ：https://leetcode-cn.com/problems/add-binary Copyright belongs to the network . For commercial reprint, please contact the official authority , Non-commercial reprint please indicate the source .

package cn.fansunion.leecode.computer; /**  * Binary sum Here are two binary strings , Returns the sum of them （ In binary ）.  *  * Input is Non empty String and contains only numbers 1 and 0.  Power button  *  * and PlusOne This question , The idea is basically the same ,PlusOne It's a variable + Fixed number 1.  *   Power button  *  * @author [email protected]  *  *         2022-2-18  */ public class AddBinary {     /* Example 1:                Input : a = "11", b = "1"          Output : "100"          Example 2:                Input : a = "1010", b = "1011"          Output : "10101"                           Tips ：                Each string consists only of characters '0' or '1' form .         1 <= a.length, b.length <= 10^4          If the string is not "0" , No leading zeros .     */     /**      * get 2 The maximum length of a string , Add one by one from low to high . Maintain the latest carry . If, in the end 1 Carry , Need to add extra      *      * @param a      * @param b      * @return      */     public String addBinary(String a, String b) {         final int lengthA = a.length();         final int lengthB = b.length();         int length = Math.max(lengthA, lengthB);         int plus = 0;         StringBuilder sb = new StringBuilder();         for (int index = 0; index < length; index++) {             int chAInt = 0;             int chBInt = 0;             if (index < lengthA) {                char chA = a.charAt(lengthA - index - 1);                //int Type conversion , Cast failed (Ascii)                //final int chAInt = (int)chA;                //final int chBInt = (int)chB;                chAInt = Character.digit(chA, 2);             }             if (index < lengthB) {                 char chB = b.charAt(lengthB - index - 1);                 chBInt = Character.digit(chB, 2);             }             int sum = chAInt + chBInt + plus;             sb.append(sum % 2);             if (sum >= 2) {                 plus = 1;             } else {                 plus = 0;             }         }         // New generation 1 The highest position         if (plus == 1) {             sb.append(plus);         }         // The low position is maintained first , Display time , High in the former         return sb.reverse().toString();     } }

Execution results ： adopt According to the details Add notes Execution time ：1 ms, In all Java Defeated in submission 100.00% Users of Memory consumption ：40.1 MB, In all Java Defeated in submission 11.10% Users of Pass the test case ：294 / 294 Show off :

package test.leecode.computer; import org.junit.Assert; import org.junit.Test; import cn.fansunion.leecode.computer.AddBinary; /**  * @author [email protected]  *  * 2022-2-18  */ public class AddBinaryTest {     @Test     public void test() {         AddBinary addBinary = new AddBinary();         Assert.assertEquals("11",addBinary.addBinary("10", "1"));         Assert.assertEquals("100",addBinary.addBinary("1", "11"));         Assert.assertEquals("110",addBinary.addBinary("11", "11"));         Assert.assertEquals("1110",addBinary.addBinary("111", "111"));         Assert.assertEquals("10101",addBinary.addBinary("1010", "1011"));         Assert.assertEquals("1111",addBinary.addBinary("1000", "111"));          } }