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| 1 | +""" |
| 2 | +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.) |
| 3 | + |
| 4 | +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'. |
| 5 | + |
| 6 | +Return the minimum number of flips to make S monotone increasing. |
| 7 | + |
| 8 | + |
| 9 | + |
| 10 | +Example 1: |
| 11 | + |
| 12 | +Input: "00110" |
| 13 | +Output: 1 |
| 14 | +Explanation: We flip the last digit to get 00111. |
| 15 | +Example 2: |
| 16 | + |
| 17 | +Input: "010110" |
| 18 | +Output: 2 |
| 19 | +Explanation: We flip to get 011111, or alternatively 000111. |
| 20 | +Example 3: |
| 21 | + |
| 22 | +Input: "00011000" |
| 23 | +Output: 2 |
| 24 | +Explanation: We flip to get 00000000. |
| 25 | + |
| 26 | + |
| 27 | +Note: |
| 28 | + |
| 29 | +1 <= S.length <= 20000 |
| 30 | +S only consists of '0' and '1' characters. |
| 31 | + |
| 32 | + |
| 33 | +给定一个由 '0' 和 '1' 组成的字符串,把它变成单调递增的字符串。返回最少的改变次数。 |
| 34 | + |
| 35 | +思路: |
| 36 | +从左向右走过一遍,把找到的 1 变成 0。 |
| 37 | +从右向左过一遍,把找到的 0 变成 1。 |
| 38 | + |
| 39 | +最后过一遍,找到相加最少的一个点即可(可能不止一个)。 |
| 40 | + |
| 41 | +测试地址: |
| 42 | +https://leetcode.com/contest/weekly-contest-107/problems/flip-string-to-monotone-increasing/ |
| 43 | + |
| 44 | +""" |
| 45 | + |
| 46 | +class Solution(object): |
| 47 | + def minFlipsMonoIncr(self, S): |
| 48 | + """ |
| 49 | + :type S: str |
| 50 | + :rtype: int |
| 51 | + """ |
| 52 | + |
| 53 | + x = [0] if S[0] == '0' else [1] |
| 54 | + |
| 55 | + # left to right |
| 56 | + # 1 -> 0 |
| 57 | + for i in range(1, len(S)): |
| 58 | + if S[i] == '1': |
| 59 | + x.append(x[-1]+1) |
| 60 | + else: |
| 61 | + x.append(x[-1]) |
| 62 | + |
| 63 | + # right to left |
| 64 | + # 0 -> 1 |
| 65 | + S = S[::-1] |
| 66 | + y = [0] if S[0] == '1' else [1] |
| 67 | + for i in range(1, len(S)): |
| 68 | + if S[i] == '0': |
| 69 | + y.append(y[-1]+1) |
| 70 | + else: |
| 71 | + y.append(y[-1]) |
| 72 | + |
| 73 | + y.reverse() |
| 74 | + |
| 75 | + return min([i+j for i,j in zip(x,y)]) - 1 |
| 76 | + |
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