Table of Contents
一、中文版
房间中有 n 个灯泡，编号从 0 到 n-1 ，自左向右排成一行。最开始的时候，所有的灯泡都是 关 着的。
请你设法使得灯泡的开关状态和 target 描述的状态一致，其中 target[i] 等于 1 第 i 个灯泡是开着的，等于 0 意味着第 i 个灯是关着的。
有一个开关可以用于翻转灯泡的状态，翻转操作定义如下：
- 选择当前配置下的任意一个灯泡（下标为
i） - 翻转下标从
i到n-1的每个灯泡
翻转时，如果灯泡的状态为 0 就变为 1，为 1 就变为 0 。
返回达成 target 描述的状态所需的 最少 翻转次数。
示例 1：
输入：target = "10111" 输出：3 解释：初始配置 "00000". 从第 3 个灯泡（下标为 2）开始翻转 "00000" -> "00111" 从第 1 个灯泡（下标为 0）开始翻转 "00111" -> "11000" 从第 2 个灯泡（下标为 1）开始翻转 "11000" -> "10111" 至少需要翻转 3 次才能达成 target 描述的状态
示例 2：
输入：target = "101" 输出：3 解释："000" -> "111" -> "100" -> "101".
示例 3：
输入：target = "00000" 输出：0
示例 4：
输入：target = "001011101" 输出：5
提示：
1 <= target.length <= 10^5target[i] == '0'或者target[i] == '1'
二、英文版
There is a room with n bulbs, numbered from 0 to n-1, arranged in a row from left to right. Initially all the bulbs are turned off.
Your task is to obtain the configuration represented by target where target[i] is '1' if the i-th bulb is turned on and is '0' if it is turned off.
You have a switch to flip the state of the bulb, a flip operation is defined as follows:
- Choose any bulb (index
i) of your current configuration. - Flip each bulb from index
iton-1.
When any bulb is flipped it means that if it is 0 it changes to 1 and if it is 1 it changes to 0.
Return the minimum number of flips required to form target.
Example 1:
Input: target = "10111" Output: 3 Explanation: Initial configuration "00000". flip from the third bulb: "00000" -> "00111" flip from the first bulb: "00111" -> "11000" flip from the second bulb: "11000" -> "10111" We need at least 3 flip operations to form target.
Example 2:
Input: target = "101" Output: 3 Explanation: "000" -> "111" -> "100" -> "101".
Example 3:
Input: target = "00000" Output: 0
Example 4:
Input: target = "001011101" Output: 5
Constraints:
1 <= target.length <= 10^5target[i] == '0'ortarget[i] == '1'
三、My answer
class Solution:
def minFlips(self, target: str) -> int:
res = 0
n = len(target)
last_one = 0
for i in range(n):
if target[i] != last_one:
res += 1
last_one = target[i]
return res
四、解题报告
数据结构：数组
算法：遍历
实现：可以不按照题目给的例子进行翻转，将例1的翻转过程改为
"00000"--> "11111"-->"10000"--> "10111"
再结合其他例子就可以总结出翻转的方法：
从左往右遍历 target 数组，遇到第一个 1 时，将 1 及以后所有数字都翻转一次；下一次就是遇到 0 时，0及后边的数字都再翻转一次；也就是每次遇到跟之前数字不一样时，后面的数字都需要翻转。
如此，便可以统计出需要翻转的总次数。