题目如下:
Given two strings
s
andt
, your goal is to converts
intot
ink
moves or less.During the
ith
(1 <= i <= k
) move you can:
- Choose any index
j
(1-indexed) froms
, such that1 <= j <= s.length
andj
has not been chosen in any previous move, and shift the character at that indexi
times.- Do nothing.
Shifting a character means replacing it by the next letter in the alphabet (wrapping around so that
‘z‘
becomes‘a‘
). Shifting a character byi
means applying the shift operationsi
times.Remember that any index
j
can be picked at most once.Return
true
if it‘s possible to converts
intot
in no more thank
moves, otherwise returnfalse
.Example 1:
Input: s = "input", t = "ouput", k = 9 Output: true Explanation: In the 6th move, we shift ‘i‘ 6 times to get ‘o‘. And in the 7th move we shift ‘n‘ to get ‘u‘.Example 2:
Input: s = "abc", t = "bcd", k = 10 Output: false Explanation: We need to shift each character in s one time to convert it into t. We can shift ‘a‘ to ‘b‘ during the 1st move.
However, there is no way to shift the other characters in the remaining moves to obtain t from s.Example 3:
Input: s = "aab", t = "bbb", k = 27 Output: true Explanation: In the 1st move, we shift the first ‘a‘ 1 time to get ‘b‘. In the 27th move, we shift the second ‘a‘ 27 times
to get ‘b‘.Constraints:
1 <= s.length, t.length <= 10^5
0 <= k <= 10^9
s
,t
contain only lowercase English letters.
解题思路:统计出s中每个字符转成成t中对应的字符所需要的转换次数,如果需要转换i次的字符的数量一个有v个,那么需要满足 i + (v-1)*26 > k 。
代码如下:
class Solution(object): def canConvertString(self, s, t, k): """ :type s: str :type t: str :type k: int :rtype: bool """ if len(s) != len(t):return False count = [0] * 27 for (c1,c2) in zip(s,t): if c1 <= c2: count[ord(c2) - ord(c1)] += 1 else: count[26 - (ord(c1) - ord(c2))] += 1 for i,v in enumerate(count): if i > 0 and i + (v-1)*26 > k: return False return True
【leetcode】1540. Can Convert String in K Moves
原文:https://www.cnblogs.com/seyjs/p/13667720.html