1. leetcode 300 最长递增子序列
class Solution {
public:
int lengthOfLIS(vector<int>& nums) {
if (nums.size() <= 1) return nums.size();
int res = 0;
vector<int> dp(nums.size(), 1);
for (int i = 1; i < nums.size(); i++) {
for (int j = 0; j < i; j++) {
if (nums[i] > nums[j]) {
dp[i] = max(dp[i], dp[j] + 1);
}
}
if (dp[i] > res) res = dp[i];
}
return res;
}
};
2. leetcode 674 最长连续递增序列
class Solution {
public:
int findLengthOfLCIS(vector<int>& nums) {
int last = 0;
int res = 1;
for (int i = 1; i < nums.size(); i++) {
if (nums[i - 1] < nums[i]) {
res = max(res, i - last + 1);
} else {
last = i;
}
}
return res;
}
};
class Solution {
public:
int findLengthOfLCIS(vector<int>& nums) {
if (nums.size() == 0) return 0;
int result = 1;
vector<int> dp(nums.size(), 1);
for (int i = 0; i < nums.size() - 1; i++) {
if (nums[i + 1] > nums[i]) {
dp[i + 1] = dp[i] + 1;
}
if (dp[i + 1] > result) result = dp[i + 1];
}
return result;
}
};
class Solution {
public:
int findLengthOfLCIS(vector<int>& nums) {
if (nums.size() == 0) return 0;
int res = 1;
int count = 1;
for (int i = 0; i < nums.size() - 1; i++) {
if (nums[i + 1] > nums[i]) {
count++;
} else {
count = 1;
}
if (count > res) res = count;
}
return res;
}
};
3. leetcode 718 最长重复子数组
class Solution {
public:
int findLength(vector<int>& nums1, vector<int>& nums2) {
vector<vector<int>> dp(nums1.size() + 1, vector<int>(nums2.size() + 1, 0));
int result = 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;
}
if (dp[i][j] > result) result = dp[i][j];
}
}
return result;
}
};
class Solution {
public:
int findLength(vector<int>& nums1, vector<int>& nums2) {
vector<int> dp(vector<int>(nums2.size() + 1, 0));
int result = 0;
for (int i = 1; i <= nums1.size(); i++) {
for (int j = nums2.size(); j > 0; j--) {
if (nums1[i - 1] == nums2[j - 1]) {
dp[j] = dp[j - 1] + 1;
} else {
dp[j] = 0;
}
if (dp[j] > result) result = dp[j];
}
}
return result;
}
};
4. leetcode 1143 最长公共子序列
class Solution {
public:
int longestCommonSubsequence(string text1, string text2) {
vector<vector<int>> dp(text1.size() + 1, vector<int>(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 - 1][j], dp[i][j - 1]);
}
}
}
return dp[text1.size()][text2.size()];
}
};
5. leetcode 1035 不相交的线
class Solution {
public:
int maxUncrossedLines(vector<int>& nums1, vector<int>& nums2) {
vector<vector<int>> dp(nums1.size() + 1, vector<int>(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 - 1][j], dp[i][j - 1]);
}
}
}
return dp[nums1.size()][nums2.size()];
}
};
6. leetcode 53 最大子数组和
class Solution {
public:
int maxSubArray(vector<int>& nums) {
if (nums.size() == 0) return 0;
vector<int> dp(nums.size());
dp[0] = nums[0];
int res = dp[0];
for (int i = 1; i < nums.size(); i++) {
dp[i] = max(dp[i - 1] + nums[i], nums[i]);
if (dp[i] > res) res = dp[i];
}
return res;
}
};
7. leetcode 392 判断子序列
class Solution {
public:
bool isSubsequence(string s, string t) {
vector<vector<int>> dp(s.size() + 1, vector<int>(t.size() + 1, 0));
for (int i = 1; i <= s.size(); i++) {
for (int j = 1; j <= t.size(); j++) {
if (s[i - 1] == t[j - 1]) {
dp[i][j] = dp[i - 1][j - 1] + 1;
} else {
dp[i][j] = dp[i][j - 1];
}
}
}
if (dp[s.size()][t.size()] == s.size()) return true;
return false;
}
};
8. leetcode 115 不同的子序列
class Solution {
public:
int numDistinct(string s, string t) {
vector<vector<uint64_t>> dp(s.size() + 1, vector<uint64_t>(t.size() + 1));
for (int i = 0; i < s.size(); i++) dp[i][0] = 1;
for (int j = 1; j < t.size(); j++) dp[0][j] = 0;
for (int i = 1; i <= s.size(); i++) {
for (int j = 1; j <= t.size(); j++) {
if (s[i - 1] == t[j - 1]) {
dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j];
} else {
dp[i][j] = dp[i - 1][j];
}
}
}
return dp[s.size()][t.size()];
}
};
9. leetcode 583 两个字符串的删除操作
class Solution {
public:
int minDistance(string word1, string word2) {
vector<vector<int>> dp(word1.size() + 1, vector<int>(word2.size() + 1));
for (int i = 0; i <= word1.size(); i++) dp[i][0] = i;
for (int j = 0; j <= word2.size(); j++) dp[0][j] = j;
for (int i = 1; i <= word1.size(); i++) {
for (int j = 1; j <= word2.size(); j++) {
if (word1[i - 1] == word2[j - 1]) {
dp[i][j] = dp[i - 1][j - 1];
} else {
dp[i][j] = min({dp[i - 1][j - 1] + 2, dp[i - 1][j] + 1, dp[i][j - 1] + 1});
}
}
}
return dp[word1.size()][word2.size()];
}
};
10. leetcode 72 编辑距离
class Solution {
public:
int minDistance(string word1, string word2) {
vector<vector<int>> dp(word1.size() + 1, vector<int>(word2.size() + 1, 0));
for (int i = 0; i <= word1.size(); i++) dp[i][0] = i;
for (int j = 0; j <= word2.size(); j++) dp[0][j] = j;
for (int i = 1; i <= word1.size(); i++) {
for (int j = 1; j <= word2.size(); j++) {
if (word1[i - 1] == word2[j - 1]) {
dp[i][j] = dp[i - 1][j - 1];
} else {
dp[i][j] = min({dp[i - 1][j - 1], dp[i - 1][j], dp[i][j - 1]}) + 1;
}
}
}
return dp[word1.size()][word2.size()];
}
};
11. leetcode 647 回文子串
class Solution {
public:
int countSubstrings(string s) {
vector<vector<bool>> dp(s.size(), vector<bool>(s.size(), false));
int result = 0;
for (int i = s.size() - 1; i >= 0; i--) {
for(int j = i; j < s.size(); j++) {
if (s[i] == s[j]) {
if (j - i <= 1) {
result++;
dp[i][j] = true;
} else if (dp[i + 1][j - 1]) {
result++;
dp[i][j] = true;
}
}
}
}
return result;
}
};
class Solution {
public:
int countSubstrings(string s) {
int result = 0;
for (int i =0; i < s.size(); i++) {
result += extend(s, i, i, s.size());
result += extend(s, i, i + 1, s.size());
}
return result;
}
int extend(const string& s, int i, int j, int n) {
int res = 0;
while (i >= 0 && j < n && s[i] == s[j]) {
i--;
j++;
res++;
}
return res;
}
};
12. leetcode 516 最长回文子序列
class Solution {
public:
int longestPalindromeSubseq(string s) {
vector<vector<int>> dp(s.size(), vector<int>(s.size(), 0));
for (int i = 0; i < s.size(); i++) dp[i][i] = 1;
for (int i = s.size() - 1; i >= 0; i--) {
for (int j = i + 1; j < s.size(); j++) {
if (s[i] == s[j]) {
dp[i][j] = dp[i + 1][j - 1] + 2;
} else {
dp[i][j] = max(dp[i + 1][j], dp[i][j - 1]);
}
}
}
return dp[0][s.size() - 1];
}
};