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'
将二进制表示减到 1 的步骤数。
给你一个以二进制形式表示的数字 s 。请你返回按下述规则将其减少到 1 所需要的步骤数:
如果当前数字为偶数,则将其除以 2 。
如果当前数字为奇数,则将其加上 1 。
题目保证你总是可以按上述规则将测试用例变为 1 。
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/number-of-steps-to-reduce-a-number-in-binary-representation-to-one
著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。
既然处理的是一个以字符串表示的二进制数字,那么我们就按照规则和需求进行处理。最终的目的是把这个二进制数变成1,所以我们从右往左扫描input字符串,如果遇到0则可以除以2,这里除以2相当于是位运算里的右移一位,需要一步;如果遇到1,则对其 + 1并且再除以2,这里需要两步。在计算的过程中我们需要记录进位carry。
此时我们有一个已经用到的步数step和一个进位carry。
如果当前位置上是0且carry = 0,则可以除以2,这里花费一步
如果当前位置上是0且carry = 1,则需要花费两步,一步是+1,另一步是除以2;此时carry仍然是1
如果当前位置上是1且carry = 0,则需要花费两步,一步是+1,另一步是除以2;此时carry是1
如果当前位置上是1且carry = 1,则可以除以2,这里花费一步
注意题目条件有这一条,s[0] == '1',所以第一位要特判,从右往左扫描的时候要在第一位停下。最后看一下,如果依然进位carry = 0,则已经满足题意;如果进位carry = 1,则加上进位之后s[0] = 0,此时要变回1则还需要再 + 1。
时间O(n)
空间O(1)
Java实现
1 class Solution { 2 public int numSteps(String s) { 3 int step = 0; 4 // corner case 5 if (s.length() == 1) { 6 if (s.charAt(0) == '0') { 7 step++; 8 } 9 return step; 10 } 11 12 // normal case 13 int carry = 0; 14 for (int i = s.length() - 1; i > 0; i--) { 15 // '0' 16 if (s.charAt(i) == '0') { 17 if (carry == 0) { 18 // 右移一步 19 step++; 20 } else { 21 step += 2; 22 carry = 1; 23 } 24 } 25 // '1' 26 else { 27 if (carry == 0) { 28 step += 2; 29 } else { 30 step++; 31 } 32 carry = 1; 33 } 34 } 35 return step + carry; 36 } 37 }