1404. Number of Steps to Reduce a Number in Binary Representation to One
作者:互联网
Given a number s
in their binary representation. Return the number of steps to reduce it to 1 under the following rules:
-
If the current number is even, you have to divide it by 2.
-
If the current number is odd, you have to add 1 to it.
It's guaranteed that you can always reach to one for all testcases.
Example 1:
Input: s = "1101" Output: 6 Explanation: "1101" corressponds to number 13 in their decimal representation. Step 1) 13 is odd, add 1 and obtain 14. Step 2) 14 is even, divide by 2 and obtain 7. Step 3) 7 is odd, add 1 and obtain 8. Step 4) 8 is even, divide by 2 and obtain 4. Step 5) 4 is even, divide by 2 and obtain 2. Step 6) 2 is even, divide by 2 and obtain 1.
Example 2:
Input: s = "10" Output: 1 Explanation: "10" corressponds to number 2 in their decimal representation. Step 1) 2 is even, divide by 2 and obtain 1.
Example 3:
Input: s = "1" Output: 0
Constraints:
1 <= s.length <= 500
s
consists of characters '0' or '1's[0] == '1'
class Solution { public int numSteps(String s) { int n = s.length(), res = 0, i = n - 1, carry = 0; char[] arr = s.toCharArray(); while (i > 0) { if (carry == 1 && arr[i] == '0' || carry == 0 && arr[i] == '1') { res += 2; carry = 1; } else if (arr[i] == '0' && carry == 0) { res++; } else { res++; carry = 1; } i--; } if (carry > 0) res++; return res; } }
public int numSteps(String s) { int n = s.length(), res = 0, i = n - 1, carry = 0; while (i > 0) { if (s.charAt(i--) - '0' + carry == 1) { // curr digit + carry is '1'; need to extra adding '1' operation; res++; carry = 1; } res++; // dividing 2; } return res + carry; // first digit must be 1, after add carry, need extra dividing 2; }
从后往前整,如果当前位 + carry位 == 0则仅需要右移一位(除以2),此时res++
如果当前位 + carry位 == 1, 则先需要+1,而且carry位此时变成1,res +=2;
如果最后剩下一位那必然是1,如果此时carry位还是1,就要再右移一位,此时res++
标签:Binary,++,res,Reduce,number,Number,Step,carry,obtain 来源: https://www.cnblogs.com/wentiliangkaihua/p/13091960.html