Code & Fun

第23天。

今天的题目是 Longest Common Subsequence :


Given two strings text1 and text2, return the length of their longest common subsequence.

A subsequence of a string is a new string generated from the original string with some characters(can be none) deleted without changing the relative order of the remaining characters. (eg, “ace” is a subsequence of “abcde” while “aec” is not). A common subsequence of two strings is a subsequence that is common to both strings.

If there is no common subsequence, return 0.

Example 1:

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Input: text1 = "abcde", text2 = "ace" 
Output: 3
Explanation: The longest common subsequence is "ace" and its length is 3.

Example 2:

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Input: text1 = "abc", text2 = "abc"
Output: 3
Explanation: The longest common subsequence is "abc" and its length is 3.

Example 3:

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Input: text1 = "abc", text2 = "def"
Output: 0
Explanation: There is no such common subsequence, so the result is 0.

Constraints:

  • 1 <= text1.length <= 1000
  • 1 <= text2.length <= 1000
  • The input strings consist of lowercase English characters only.

这是一道比较经典的动态规划问题吧,它的动规方程为:
$$
\begin{equation}
LCS(i,j) = \left{
\begin{array}{rcl}

& LCS[i-1, j-1] + 1 & ,{s1[i] = s2[j]} \
& max(LCS[i, j-1], LCS[i-1, j]) & ,{s1[i] \neq s2[j]}

\end{array}
\right.
\end{equation}
$$
根据动规方程我们可以写出如下代码:

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int longestCommonSubsequence(string text1, string text2) {
vector<vector<int>> dp(text1.size() + 1, vector<int>(text2.size() + 1, 0));
for(int i = 1;i < dp.size(); i++) {
for(int j = 1;j < dp[0].size(); j++) {
if (text1[i-1] == text2[j-1]) {
dp[i][j] = dp[i-1][j-1] + 1;
} else {
dp[i][j] = max(dp[i-1][j], dp[i][j-1]);
}
}
}
return dp[text1.size()][text2.size()];
}

这里的空间复杂度可以继续进行优化,因为LCS[i,j]只与当前行和上一行有关系,所以可以优化成两个数组来做:

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int longestCommonSubsequence(string text1, string text2) {

int n = text1.size() + 1, m = text2.size() + 1;
vector<int> dp1(m, 0);
vector<int> dp2(m);

for(int i = 1;i < n; i++) {
dp1[0] = 0;
for(int j = 1;j < m; j++) {
if (text1[i-1] == text2[j-1]) {
dp2[j] = dp1[j-1] + 1;
} else {
dp2[j] = max(dp1[j], dp2[j-1]);
}
}
swap(dp1, dp2);
}
return dp1[m-1];
}

再进一步的话,我们可以发现dp[i][j]只与 dp[i-1][j-1]dp[i-1][j]dp[i][j-1] 相关,如果我们只用一个数组的话,dp[i][j]dp[i-1][j]其实存在同一个位置,而dp[i][j-1]是在同一行,所以我们只需要维护一个prev变量来保存dp[i-1][j-1]的值即可:

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int longestCommonSubsequence(string text1, string text2) {
int n = text1.size(), m = text2.size() + 1;
vector<int> dp(m, 0);

for(int i = 0;i < n; i++) {
int prev = 0;
for(int j = 1;j < m; j++) {
int temp = prev;
prev = dp[j];
dp[j] = (text1[i] == text2[j-1]) ? (temp + 1) : (max(dp[j], dp[j-1]));
}
}
return dp[m-1];
}

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