# 力扣第44天----第1143题、第1035题
上一题是,dp[i][j] 与 dp[i-1][j-1]建立关系。这题是dp[i][j] 与 dp[i-1][j-1]、dp[i][j-1]、dp[i-1][j]建立递推关系。
class Solution {
public:
int longestCommonSubsequence(string text1, string text2) {
vector> dp(text1.size()+1, vector(text2.size() + 1, 0));
for(int i =1; i<=text1.size(); ++i){
for (int j =1; j<=text2.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][j-1], dp[i-1][j]);
}
}
return dp[text1.size()][text2.size()];
}
};
套了个壳子,实际跟上一题一模一样。
class Solution {
public:
int maxUncrossedLines(vector& nums1, vector& nums2) {
vector> dp(nums1.size()+1, vector(nums2.size() + 1, 0));
for(int i =1; i<=nums1.size(); ++i){
for (int j =1; j<=nums2.size(); ++j){
if(nums1[i-1] == nums2[j-1]) dp[i][j] = dp[i-1][j-1] +1;
else dp[i][j] = max(dp[i][j-1], dp[i-1][j]);
}
}
return dp[nums1.size()][nums2.size()];
}
};