1745. Palindrome Partitioning IV
Given a string s, return true if it is possible to split the string s into three non-empty palindromic substrings. Otherwise, return false.
A string is said to be palindrome if it the same string when reversed.
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| class Solution {
public:
bool checkPartitioning(string s) {
string extend = "#";
for(auto& ch : s) {
extend.push_back(ch);
extend.push_back('#');
}
int l = 0, r = -1, n = extend.length();
vector<int> dp(n);
vector<int> bound;
for(int i = 0; i < n; i++) {
dp[i] = max(0, min(r - i, (r + l - i >= 0? dp[r + l - i] : -1)));
while(i + dp[i] < n and i - dp[i] >= 0 and extend[i + dp[i]] == extend[i - dp[i]])
dp[i]++;
if(i > 0 and i - dp[i] < 0)
bound.push_back(i + dp[i]);
if(i + dp[i] > r) {
r = i + dp[i];
l = i - dp[i];
}
}
sort(begin(bound), end(bound));
for(int i = n - 2; i >= 0; i--) {
if(i + dp[i] < n) continue;
int left = i - dp[i];
for(int i = 0; i < bound.size() and bound[i] <= left; i++) {
int mid = (bound[i] + left) / 2;
int dis = left - mid;
if(bound[i] == left and extend[mid] == '#') continue;
if(dp[mid] > left - mid) return true;
}
}
return false;
}
};
|
