A string of ‘0‘
s and ‘1‘
s is monotone increasing if it consists of some number of ‘0‘
s (possibly 0),followed by some number of ‘1‘
s (also possibly 0.)
We are given a string S
of ‘0‘
s and ‘1‘
s,and we may flip any ‘0‘
to a ‘1‘
or a ‘1‘
to a ‘0‘
.
Return the minimum number of flips to make S
monotone increasing.
Example 1:
Input: "00110"
Output: 1 Explanation: We flip the last digit to get 00111.
Example 2:
Input: "010110"
Output: 2 Explanation: We flip to get 011111,or alternatively 000111.
Example 3:
Input: "00011000"
Output: 2 Explanation: We flip to get 00000000.
Note:
1 <= S.length <= 20000
-
S
only consists of‘0‘
and‘1‘
characters.
如果一个由 ‘0‘
和 ‘1‘
组成的字符串,是以一些 ‘0‘
(可能没有 ‘0‘
)后面跟着一些 ‘1‘
(也可能没有 ‘1‘
)的形式组成的,那么该字符串是单调递增的。
我们给出一个由字符 ‘0‘
和 ‘1‘
组成的字符串 S
,我们可以将任何 ‘0‘
翻转为 ‘1‘
或者将 ‘1‘
翻转为 ‘0‘
。
返回使 S
单调递增的最小翻转次数。
示例 1:
输入:"00110" 输出:1 解释:我们翻转最后一位得到 00111.
示例 2:
输入:"010110" 输出:2 解释:我们翻转得到 011111,或者是 000111。
示例 3:
输入:"00011000" 输出:2 解释:我们翻转得到 00000000。
提示:
1 <= S.length <= 20000
-
S
中只包含字符‘0‘
和‘1‘
116ms
1 class Solution { 2 func minFlipsMonoIncr(_ S: String) -> Int { 3 var arr:[Character] = [Character]() 4 for char in S.characters 5 { 6 arr.append(char) 7 } 8 let len = S.count 9 //声明数组 10 var ct:[Int] = [Int](repeating: 0,count: len + 1) 11 var x:Int = 0 12 for i in 0...len 13 { 14 ct[i] += x 15 if i < len && arr[i] == "1" 16 { 17 x += 1 18 } 19 } 20 x = 0 21 for i in (0...len).reversed() 22 { 23 ct[i] += x 24 if i > 0 && arr[i - 1] == "0" 25 { 26 x += 1 27 } 28 } 29 var res = 999999999 30 for i in 0...len 31 { 32 res = min(res,ct[i]) 33 } 34 return res 35 } 36 }