Programming
Leetcode-67. Add Binary
Given two binary strings a and b, return their sum as a binary string.
Example 1:
Input: a = "11", b = "1"
Output: "100"
Example 2:
Input: a = "1010", b = "1011"
Output: "10101"
<解題>
- 用carry保存進位的值，由後面往前算，數字的值相加進carry中
- while (i >= 0 || j >= 0|| carry==1)，代表還有i,j數字要處理或者還有進位要處理
- sb.append(carry % 2); //當前位子的結果0或1
- carry /= 2做進位的處理
- 最後把這個sb反轉字串返回
public class Solution {
public String addBinary(String a, String b) {
StringBuilder sb = new StringBuilder();
int i = a.length() - 1, j = b.length() -1, carry = 0; //保存進位的值
while (i >= 0 || j >= 0|| carry==1) {
if (j >= 0) {
carry += b.charAt(j--) - '0';
}
if (i >= 0) {
carry += a.charAt(i--) - '0';
}
sb.append(carry % 2); //當前位子的結果0,1
carry /= 2; //carry/2 以進行進位處理
}
return sb.reverse().toString();
}
}
Time: O(max(m, n))->兩字串長度
Space: O(max(m, n)
另外一種寫法
public class Solution {
public String addBinary(String a, String b) {
StringBuilder sb = new StringBuilder();
int i = a.length() - 1, j = b.length() -1, carry = 0;
// a = "1010", b = "1011"
while (i >= 0 || j >= 0|| carry==1) {
if (i >= 0) {
carry += a.charAt(i) - '0';
}
if (j >= 0) {
carry += b.charAt(j) - '0';
}
sb.append(carry % 2); //當前位子的結果0,1
carry /= 2; //carry/2 以進行進位處理
i--;
j--;
}
return sb.reverse().toString();
}
}