leetcode:5. Longest Palindromic Substring

描述

Given a string s, find the longest palindromic substring in s. You may assume that the maximum length of s is 1000.

Example:

Input: “babad”

Output: “bab”

Note: “aba” is also a valid answer.

Example:

Input: “cbbd”

Output: “bb”

思路

遍历字符串s, 遍历索引cur，检测cur两边是不是相等。

代码

class Solution {
public:
string longestPalindrome(string s) {
int size_s = s.size();
int cur = 0, sub_cur = 0,max_palindromic = 1, start_idx = 0;
bool is_same = true;
while(cur < size_s){
if(s[cur+1] == s[cur]){
sub_cur = 1;
is_same = true;
while(is_same ){
is_same = (s[cur - sub_cur] == s[cur + sub_cur + 1] && cur - sub_cur>=0) ? true:false;
sub_cur++;
}
if(2*sub_cur - 2> max_palindromic){
start_idx = cur - sub_cur + 2;
max_palindromic = 2*sub_cur - 2;
}
}
if(s[cur-1] == s[cur + 1] && cur - 1 >= 0){
sub_cur = 2;
is_same = true;
while(is_same ){
is_same = (s[cur - sub_cur] == s[cur + sub_cur] && cur - sub_cur >=0) ? true:false;
sub_cur++;
}
if(2*sub_cur - 3> max_palindromic){
start_idx = cur - sub_cur + 2;
max_palindromic = 2*sub_cur - 3;
}
}
cur++;
}

return s.substr(start_idx, max_palindromic);
}
};

结果

反思

今天原计划刷两题的，但是这一题耽误了我太多时间，大概40-50分钟的时候就已经编写出来了，但是一直在找错，最后发现是string的一个接口（substr）没有理解清楚。！！！！谨记，使用别人接口的时候，一定要弄明白参数的每一个含义。

他山之玉

C++ O(n) solutioin

string longestPalindrome(string s) {
if (s.empty()) return "";
if (s.size() == 1) return s;
int min_start = 0, max_len = 1;
for (int i = 0; i < s.size();) {
if (s.size() - i <= max_len / 2) break;
int j = i, k = i;
while (k < s.size()-1 && s[k+1] == s[k]) ++k; // Skip duplicate characters.
i = k+1;
while (k < s.size()-1 && j > 0 && s[k + 1] == s[j - 1]) { ++k; --j; } // Expand.
int new_len = k - j + 1;
if (new_len > max_len) { min_start = j; max_len = new_len; }
}
return s.substr(min_start, max_len);
}

Java O(n) solution

public String longestPalindrome(String s) {
if(s==null){
return "";
}
char[] arr = s.toCharArray();
int max = 0;
int maxi = 0;
int maxj = 0;

for(int i = 0; i< arr.length;){
int i1 = getFarestSameElementIndex(arr,i);
int dist = getDistance(arr,i,i1);
int index1 = i-dist;
int index2 = i1 + dist;
int l = index2 - index1;
if(l>max){
max = l;
maxi = index1;
maxj = index2;
}
i = i1+1;
}

return s.substring(maxi, maxj+1);
}

private int getDistance(char[] arr,int index1,int index2){
int i1 = index1-1;
int i2 = index2+1;
int dist = 0;
while(i1>=0&&i2<arr.length){
if(arr[i1]==arr[i2]){
dist++;
}else{
break;
}
i1--;i2++;
}
return dist;
}

private int getFarestSameElementIndex(char[] arr, int index){
for(int i = index+1;i<arr.length;i++){
if(arr[i]!=arr[index]){
return i-1;
}
}
return arr.length-1;
}

python O(n) solution

lass Solution(object):
def longestPalindrome(self, s):
"""
:type s: str
:rtype: str
"""
# make it okay for both odd and even cases
T = "^#" + "#".join(list(s)) + "#$"
n = len(T)
P = [0] * n
# C and R are center and right boundary of current farest palindrome
C = R = 0
for i in range(1, n-1):
P[i] = min(R-i, P[2*C-i]) if R > i else 0
# check palindrome only for possible candidate, key idea why it's O(n)
if P[i] + i >= R:
while T[i-1-P[i]] == T[i+1+P[i]]:
P[i] += 1
# update C and R
C, R = i, i + P[i]
L, C = max((x, j) for j, x in enumerate(P))
return s[(C - L) >> 1 : (C + L) >> 1]