diff --git a/Dynamic_Programming/2472.Maximum-Number-of-Non-overlapping-Palindrome-Substrings/2472.Maximum-Number-of-Non-overlapping-Palindrome-Substrings.cpp b/Dynamic_Programming/2472.Maximum-Number-of-Non-overlapping-Palindrome-Substrings/2472.Maximum-Number-of-Non-overlapping-Palindrome-Substrings.cpp
new file mode 100644
index 000000000..e35670f24
--- /dev/null
+++ b/Dynamic_Programming/2472.Maximum-Number-of-Non-overlapping-Palindrome-Substrings/2472.Maximum-Number-of-Non-overlapping-Palindrome-Substrings.cpp
@@ -0,0 +1,32 @@
+class Solution {
+ int isPalin[2001][2001];
+public:
+ int maxPalindromes(string s, int k) 
+ {
+ int n = s.size();
+ for (int i=0; i<n; i++) + isPalin[i][i] = 1; + for (int i=0; i+1<n; i++) + isPalin[i][i+1] = (s[i]==s[i+1]); + + for (int len=3; len<=n; len++) + for (int i=0; i+len-1<n; i++) + { + int j = i+len-1; + if (s[i]==s[j]) + isPalin[i][j] = isPalin[i+1][j-1]; + } + + vector<int>dp(n); 
+ for (int i=k-1; i<n; i++) + { + dp[i] = i==0?0:dp[i-1]; + for (int j=0; j<=i-k+1; j++) + { + if (isPalin[j][i]) + dp[i] = max(dp[i], (j==0?0:dp[j-1])+1); + } + } + return dp[n-1]; + } +}; diff --git a/Dynamic_Programming/2472.Maximum-Number-of-Non-overlapping-Palindrome-Substrings/Readme.md b/Dynamic_Programming/2472.Maximum-Number-of-Non-overlapping-Palindrome-Substrings/Readme.md new file mode 100644 index 000000000..dfb750a22 --- /dev/null +++ b/Dynamic_Programming/2472.Maximum-Number-of-Non-overlapping-Palindrome-Substrings/Readme.md @@ -0,0 +1,5 @@ +### 2472.Maximum-Number-of-Non-overlapping-Palindrome-Substrings + +令dp[i]表示s[0:i]里面的the maximum number of palindrome substrings. 显然,转移方程的突破口在于最后一个回文子串。如果最后一个回文子串不包括s[i],那么有`dp[i] = dp[i-1]`,如果最后一个回文子串包括s[i],那么我们就需要找这个回文子串的起始位置j,然后`dp[i] = dp[j-1] + 1`. 这样的j可能有多个,根据数据量,从[0:i]遍历一遍都是可行的。 + +另外,我们需要用o(N^2)的时间提前处理得到一个数组isPalin[i][j],来记录s[i:j]是否是一个回文串。这个技巧已经出现过很多次了。 diff --git a/Readme.md b/Readme.md index 509d25d92..81de381d6 100644 --- a/Readme.md +++ b/Readme.md @@ -32,6 +32,7 @@ [1798.Maximum-Number-of-Consecutive-Values-You-Can-Make](https://github.com/wisdompeak/LeetCode/blob/master/Greedy/1798.Maximum-Number-of-Consecutive-Values-You-Can-Make) (H-) [1989.Maximum-Number-of-People-That-Can-Be-Caught-in-Tag](https://github.com/wisdompeak/LeetCode/tree/master/Two_Pointers/1989.Maximum-Number-of-People-That-Can-Be-Caught-in-Tag) (M+) [2354.Number-of-Excellent-Pairs](https://github.com/wisdompeak/LeetCode/tree/master/Two_Pointers/2354.Number-of-Excellent-Pairs) (H-) +[2422.Merge-Operations-to-Turn-Array-Into-a-Palindrome](https://github.com/wisdompeak/LeetCode/tree/master/Two_Pointers/2422.Merge-Operations-to-Turn-Array-Into-a-Palindrome) (H-) * ``Sliding window`` [532.K-diff-Pairs-in-an-Array](https://github.com/wisdompeak/LeetCode/tree/master/Two_Pointers/532.K-diff-Pairs-in-an-Array) (H-) [611.Valid-Triangle-Number](https://github.com/wisdompeak/LeetCode/tree/master/Two_Pointers/611.Valid-Triangle-Number) (M+) @@ -743,6 +744,7 @@ [1478.Allocate-Mailboxes](https://github.com/wisdompeak/LeetCode/tree/master/Dynamic_Programming/1478.Allocate-Mailboxes) (H) [1977.Number-of-Ways-to-Separate-Numbers](https://github.com/wisdompeak/LeetCode/tree/master/Dynamic_Programming/1977.Number-of-Ways-to-Separate-Numbers) (H) [2463.Minimum-Total-Distance-Traveled](https://github.com/wisdompeak/LeetCode/tree/master/Dynamic_Programming/2463.Minimum-Total-Distance-Traveled) (M+) +[2472.Maximum-Number-of-Non-overlapping-Palindrome-Substrings](https://github.com/wisdompeak/LeetCode/tree/master/Dynamic_Programming/2472.Maximum-Number-of-Non-overlapping-Palindrome-Substrings) (M+) * ``区间型 II`` [131.Palindrome-Partitioning](https://github.com/wisdompeak/LeetCode/tree/master/Dynamic_Programming/131.Palindrome-Partitioning) (M+) [312.Burst-Balloons](https://github.com/wisdompeak/LeetCode/tree/master/DFS/312.Burst-Balloons) (H-) diff --git a/Two_Pointers/2422.Merge-Operations-to-Turn-Array-Into-a-Palindrome/2422.Merge-Operations-to-Turn-Array-Into-a-Palindrome.cpp b/Two_Pointers/2422.Merge-Operations-to-Turn-Array-Into-a-Palindrome/2422.Merge-Operations-to-Turn-Array-Into-a-Palindrome.cpp new file mode 100644 index 000000000..a2b94b71a --- /dev/null +++ b/Two_Pointers/2422.Merge-Operations-to-Turn-Array-Into-a-Palindrome/2422.Merge-Operations-to-Turn-Array-Into-a-Palindrome.cpp @@ -0,0 +1,35 @@ +using LL = long long; +class Solution { +public: + int minimumOperations(vector<int>& nums) 
+ {
+ int i = 0, j = nums.size()-1;
+ LL left = nums[i], right = nums[j]; 
+ int count = 0;
+ 
+ while (i<j) + { + if (left==right) + { + i++; + j--; + left = nums[i]; + right = nums[j]; + } + else if (left < right) + { + i++; + left += nums[i]; + count++; + } + else if (left> right)
+ {
+ j--;
+ right += nums[j];
+ count++;
+ }
+ }
+ return count;
+ 
+ }
+};
diff --git a/Two_Pointers/2422.Merge-Operations-to-Turn-Array-Into-a-Palindrome/Readme.md b/Two_Pointers/2422.Merge-Operations-to-Turn-Array-Into-a-Palindrome/Readme.md
new file mode 100644
index 000000000..7ff199e20
--- /dev/null
+++ b/Two_Pointers/2422.Merge-Operations-to-Turn-Array-Into-a-Palindrome/Readme.md
@@ -0,0 +1,7 @@
+### 2422.Merge-Operations-to-Turn-Array-Into-a-Palindrome
+
+我们可以设想,最终得到的回文串的最左边和最右边的元素是怎么得到的?一定是来自nums最左侧的若干个元素之和,与nums最右侧的若干个元素之和。考虑到所有的元素都是正数,我们可以用双指针的方法来得到two equal sum。也就是说,初始令`left=nums[0]`和`right=nums[n-1]`,如果发现`left<right`,必然只能将左指针右移才能试图让left与right相等;同理,如果发现`left>right`,必然只能将右指针左移才能试图让left与right相等。于是在得到`left==right`之前,两侧指针移动的总次数就是merge的次数。此时,说明我们找到了最终回文串的最外层的一对。剥离掉这最外层后,我们可以重复上述的过程。
+
+此外,我们必须考虑到,有可能在左右指针相遇之前,都无法满足`left==right`。这意味着什么呢?说明只有一个方案,即将整个数组都归并在一起,成为回文串的中心。
+
+总结:所以本题就是一个双指针,不停地调整左右指针,试图使得前缀和等于后缀和。如此一轮一轮地确定外层的每一对。如果最终指针相遇,意味着该轮的所有元素必须都归并在一起。
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