题目：
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'
代码如下：
使用flag={‘0’，‘1’}来标记当前位及当前位后面的位的状态，如果相等，继续比较下一位，不相等，反转并计数加1
class Solution {
public:int minFlips(string target) {int len = target.length();string flag({'0','1'});int cnt = 0;int ires = 0;for(int i = 0; i < len; i++){if(target[i] == flag[cnt]){continue;}else{ires++;cnt = (cnt + 1)%2;}}return ires;}
};