Interleaving String

描述

Given s1, s2, s3, find whether s3 is formed by the interleaving of s1 and s2.

For example, Given: s1 = "aabcc", s2 = "dbbca",

When s3 = "aadbbcbcac", return true.

When s3 = "aadbbbaccc", return false.

分析

设状态f[i][j]，表示s1[0,i]和s2[0,j]，匹配s3[0, i+j]。如果s1的最后一个字符等于s3的最后一个字符，则f[i][j]=f[i-1][j]；如果s2的最后一个字符等于s3的最后一个字符，则f[i][j]=f[i][j-1]。因此状态转移方程如下：

f[i][j] = (s1[i - 1] == s3 [i + j - 1] && f[i - 1][j])
       || (s2[j - 1] == s3 [i + j - 1] && f[i][j - 1]);

递归

// Interleaving String
// 递归，会超时，仅用来帮助理解
class Solution {
public:
    bool isInterleave(const string& s1, const string& s2, const string& s3) {
        if (s3.length() != s1.length() + s2.length())
            return false;

        return isInterleave(begin(s1), end(s1), begin(s2), end(s2),
                begin(s3), end(s3));
    }

    template
    bool isInterleave(InIt first1, InIt last1, InIt first2, InIt last2,
            InIt first3, InIt last3) {
        if (first3 == last3)
            return first1 == last1 && first2 == last2;

        return (*first1 == *first3
                && isInterleave(next(first1), last1, first2, last2,
                        next(first3), last3))
                || (*first2 == *first3
                        && isInterleave(first1, last1, next(first2), last2,
                                next(first3), last3));
    }
};

动规

// Interleaving String
// 二维动规，时间复杂度O(n^2)，空间复杂度O(n^2)
class Solution {
public:
    bool isInterleave(const string& s1, const string& s2, const string& s3) {
        if (s3.length() != s1.length() + s2.length())
            return false;

        vector> f(s1.length() + 1,
                vector(s2.length() + 1, true));

        for (size_t i = 1; i 

动规+滚动数组

// Interleaving String
// 二维动规+滚动数组，时间复杂度O(n^2)，空间复杂度O(n)
class Solution {
public:
    bool isInterleave(const string& s1, const string& s2, const string& s3) {
        if (s1.length() + s2.length() != s3.length())
            return false;

        if (s1.length() < s2.length())
            return isInterleave(s2, s1, s3);

        vector f(s2.length() + 1, true);

        for (size_t i = 1; i