★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
➤微信公众号:为敢(WeiGanTechnologies)
➤博客园地址:山青咏芝(https://www.cnblogs.com/strengthen/)
➤GitHub地址:https://github.com/strengthen/LeetCode
➤原文地址:https://www.cnblogs.com/strengthen/p/11443477.html
➤如果链接不是山青咏芝的博客园地址,则可能是爬取作者的文章。
➤原文已修改更新!强烈建议点击原文地址阅读!支持作者!支持原创!
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
Given a string s
, we make queries on substrings of s
.
For each query queries[i] = [left, right, k]
, we may rearrange the substring s[left], ..., s[right]
, and then choose up to k
of them to replace with any lowercase English letter.
If the substring is possible to be a palindrome string after the operations above, the result of the query is true
. Otherwise, the result is false
.
Return an array answer[]
, where answer[i]
is the result of the i
-th query queries[i]
.
Note that: Each letter is counted individually for replacement so if for example s[left..right] = "aaa"
, and k = 2
, we can only replace two of the letters. (Also, note that the initial string s
is never modified by any query.)
Example :
Input: s = "abcda", queries = [[3,3,0],[1,2,0],[0,3,1],[0,3,2],[0,4,1]] Output: [true,false,false,true,true] Explanation: queries[0] : substring = "d", is palidrome. queries[1] : substring = "bc", is not palidrome. queries[2] : substring = "abcd", is not palidrome after replacing only 1 character. queries[3] : substring = "abcd", could be changed to "abba" which is palidrome. Also this can be changed to "baab" first rearrange it "bacd" then replace "cd" with "ab". queries[4] : substring = "abcda", could be changed to "abcba" which is palidrome.
Constraints:
1 <= s.length, queries.length <= 10^5
0 <= queries[i][0] <= queries[i][1] < s.length
0 <= queries[i][2] <= s.length
s
only contains lowercase English letters.
给你一个字符串 s
,请你对 s
的子串进行检测。
每次检测,待检子串都可以表示为 queries[i] = [left, right, k]
。我们可以 重新排列 子串 s[left], ..., s[right]
,并从中选择 最多 k
项替换成任何小写英文字母。
如果在上述检测过程中,子串可以变成回文形式的字符串,那么检测结果为 true
,否则结果为 false
。
返回答案数组 answer[]
,其中 answer[i]
是第 i
个待检子串 queries[i]
的检测结果。
注意:在替换时,子串中的每个字母都必须作为 独立的 项进行计数,也就是说,如果 s[left..right] = "aaa"
且 k = 2
,我们只能替换其中的两个字母。(另外,任何检测都不会修改原始字符串 s
,可以认为每次检测都是独立的)
示例:
输入:s = "abcda", queries = [[3,3,0],[1,2,0],[0,3,1],[0,3,2],[0,4,1]] 输出:[true,false,false,true,true] 解释: queries[0] : 子串 = "d",回文。 queries[1] : 子串 = "bc",不是回文。 queries[2] : 子串 = "abcd",只替换 1 个字符是变不成回文串的。 queries[3] : 子串 = "abcd",可以变成回文的 "abba"。 也可以变成 "baab",先重新排序变成 "bacd",然后把 "cd" 替换为 "ab"。 queries[4] : 子串 = "abcda",可以变成回文的 "abcba"。
提示:
1 <= s.length, queries.length <= 10^5
0 <= queries[i][0] <= queries[i][1] < s.length
0 <= queries[i][2] <= s.length
s
中只有小写英文字母