LeetCode Medium Longest Palindromeic Substring

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

Example 1:

Input: "babad"
Output: "bab"
Note: "aba" is also a valid answer.

Example 2:

Input: "cbbd"
Output: "bb"

Solution1

DP思想

狀態轉移方程：

dp[i][j] = dp[i+1][j-1] == 1 && s[i] == s[j] ? 1 : 0

但是要注意的是，如果按照i和j從小到大的順序來枚舉子串的兩個端點，然後更新dp的話，會無法保證dp[i+1][j-1]已經被計算過，從而無法得到正確的dp

一個解決辦法時對子串的長度進行枚舉

char* longestPalindrome(char* s) {
    int len = strlen(s);
    if (len == 0) return "";
    int i, j, L, start = 0, end = 0;
    char dp[len][len];
    memset(dp, 0, sizeof(dp));
    for (i = 0; i < len; i++) {
        dp[i][i] = 1;
        if (i < len - 1) {
            if (s[i] == s[i+1]) {
                dp[i][i+1] = 1;
                start = i;
                end = i + 1;
            }
        }
    }
    
    // 對子串長度進行遞增循環
    for (L = 3; L <= len; L++) {
        for (i = 0; i + L - 1 < len; i++) {
            j = i + L - 1;
            if (s[i] == s[j] && dp[i+1][j-1] == 1) {
                dp[i][j] = 1;
                start = i;
                end = j;
            }
        }
    }
    
    char* ans = calloc(end - start + 2, sizeof(char));
    for (i = start; i <= end; i++) {
        ans[i - start] = s[i];
    }
    
    return ans;
}

Solution2

掃描

這是一個比較巧的方法，先找到可能的迴文串的中間，然後從中間向兩頭進行擴散，複雜度爲線性，而且空間複雜度爲常數！

char* longestPalindrome(char* s) {
    int len = strlen(s);
    if (len < 2) return s;
    int start = 0, end = 0;
    int max = 0;
    int i = 0;
    while (i < len) {
        int l = i;
        int r = i;
        while (r < len - 1 && s[r] == s[r + 1]) {
            r++;
        }
        i = r + 1;
        while (l > 0 && r < len - 1 && s[l - 1] == s[r + 1]) {
            r++;
            l--;
        }
        if (r - l + 1 > max) {
            max = r - l + 1;
            start = l;
            end = r;
        }
    }
    
    
    char* ans = calloc(end - start + 2, sizeof(char));
    for (i = start; i <= end; i++) {
        ans[i - start] = s[i];
    }
    
    return ans;
}