LeetCode Hot 100：动态规划
70. 爬楼梯
cpp
class Solution {
public:
int climbStairs(int n) {
if (n == 0)
return 0;
vector<int> dp(n + 1);
// 初始化
dp[0] = 1;
// 状态转移
for (int i = 1; i <= n; i++) {
dp[i] += dp[i - 1];
if (i >= 2)
dp[i] += dp[i - 2];
}
return dp[n];
}
};118. 杨辉三角
cpp
class Solution {
public:
vector<vector<int>> generate(int numRows) {
vector<vector<int>> ans(numRows);
for (int i = 0; i < numRows; i++) {
ans[i].resize(i + 1);
ans[i][0] = ans[i][i] = 1;
}
for (int i = 1; i < numRows; i++)
for (int j = 1; j < i; j++)
ans[i][j] = ans[i - 1][j - 1] + ans[i - 1][j];
return ans;
}
};198. 打家劫舍
cpp
class Solution {
public:
int rob(vector<int>& nums) {
if (nums.empty())
return 0;
int n = nums.size();
if (n == 1)
return nums[0];
vector<int> dp(n + 1);
// 初始化
dp[0] = 0;
dp[1] = nums[0];
// 状态转移
for (int i = 2; i <= n; i++)
dp[i] = max(dp[i - 1], dp[i - 2] + nums[i - 1]);
return dp[n];
}
};279. 完全平方数
思路 1：动态规划
cpp
class Solution {
public:
int numSquares(int n) {
// dp[i]: 最少需要多少个数的平方来表示 i
vector<int> dp(n + 1);
// 初始化
dp[0] = 0;
for (int i = 1; i <= n; i++) {
int minCnt = INT_MAX;
for (int j = 1; j * j <= i; j++)
minCnt = min(minCnt, dp[i - j * j]);
dp[i] = minCnt + 1;
}
return dp[n];
}
};思路 2：记忆化搜索
py
@cache
def dfs(i: int, j: int) -> int:
if i == 0:
return inf if j else 0
if j < i * i:
return dfs(i - 1, j) # 只能不选
res1 = dfs(i - 1, j) # 不选
res2 = dfs(i, j - i * i) + 1 # 选
return min(res1, res2)
class Solution:
def numSquares(self, n: int) -> int:
return dfs(isqrt(n), n)322. 零钱兑换
cpp
class Solution {
public:
int coinChange(vector<int>& coins, int amount) {
if (coins.empty())
return -1;
if (amount == 0)
return 0;
int n = coins.size();
if (n == 1 && amount % coins[0])
return -1;
// dp[i]:
vector<int> dp(amount + 1, INT_MAX / 2);
// 初始化
dp[0] = 0;
// 状态转移
for (int i = 1; i <= amount; i++)
for (int& coin : coins) {
if (i < coin)
continue;
dp[i] = min(dp[i], dp[i - coin] + 1);
}
return dp[amount] == INT_MAX / 2 ? -1 : dp[amount];
}
};139. 单词拆分
cpp
class Solution {
public:
bool wordBreak(string s, vector<string>& wordDict) {
int n = s.length();
// dp[i]: s[0,...,i] 是否可以利用 wordDict 拼接出来
vector<bool> dp(n + 1, false);
// 初始化
dp[0] = true;
// 状态转移
for (int i = 1; i <= n; i++)
for (string& word : wordDict) {
int len = word.length();
if (i >= len && s.substr(i - len, len) == word)
dp[i] = dp[i] | dp[i - len];
}
return dp[n];
}
};300. 最长递增子序列
思路 1：贪心 + 二分查找
cpp
class Solution {
public:
int lengthOfLIS(vector<int>& nums) {
vector<int> lis;
for (int& num : nums) {
auto it = lower_bound(lis.begin(), lis.end(), num);
if (it == lis.end())
lis.push_back(num); // >=num 的 lis[j] 不存在
else
*it = num;
}
return lis.size();
}
};思路 2：动态规划
cpp
class Solution {
public:
int lengthOfLIS(vector<int>& nums) {
if (nums.empty())
return 0;
int n = nums.size();
if (n == 1)
return 1;
vector<int> dp(n, 1);
// 初始化
for (int i = 0; i < n; i++)
dp[i] = 1;
int maxLen = 0;
// 状态转移
for (int i = 0; i < n; i++)
for (int j = 0; j < i; j++) {
if (nums[j] < nums[i])
dp[i] = max(dp[i], dp[j] + 1);
maxLen = max(maxLen, dp[i]);
}
return maxLen;
}
};152. 乘积最大子数组
cpp
class Solution {
public:
int maxProduct(vector<int>& nums) {
int n = nums.size();
vector<int> minF(n, 0), maxF(n, 0);
// 初始化
minF[0] = nums[0];
maxF[0] = nums[0];
// 状态转移
for (int i = 1; i < n; i++) {
minF[i] =
min({nums[i], minF[i - 1] * nums[i], maxF[i - 1] * nums[i]});
maxF[i] =
max({nums[i], minF[i - 1] * nums[i], maxF[i - 1] * nums[i]});
}
return *max_element(maxF.begin(), maxF.end());
}
};416. 分割等和子集
思路 1：0-1背包
cpp
class Solution {
public:
bool canPartition(vector<int>& nums) {
if (nums.empty())
return false;
int sum = accumulate(nums.begin(), nums.end(), 0);
if (sum % 2 == 1)
return false;
int n = nums.size();
int target = sum / 2;
// dp[i,j]：前i个数字、总和不超过j的情况下，能否满足j == target
vector<vector<bool>> dp(n + 1, vector<bool>(target + 1, false));
// 初始化
dp[0][0] = true;
// 状态转移
for (int i = 1; i <= n; i++)
for (int j = 1; j <= target; j++) {
int w = nums[i - 1];
if (j >= w)
dp[i][j] = dp[i - 1][j] || dp[i - 1][j - w]; // 选
else
dp[i][j] = dp[i - 1][j]; // 不选
}
return dp[n][target];
}
};思路 1：栈
cpp
class Solution {
public:
int longestValidParentheses(string s) {
if (s.empty())
return 0;
int maxLen = 0;
stack<int> stk;
stk.push(-1);
for (int i = 0; i < s.length(); i++) {
if (s[i] == '(')
stk.push(i);
else {
stk.pop();
if (stk.empty())
stk.push(i);
else
maxLen = max(maxLen, i - stk.top());
}
}
return maxLen;
}
};思路 2：动态规划
cpp
class Solution {
public:
int longestValidParentheses(string s) {
if (s.empty())
return 0;
int n = s.length();
// dp[i]: 表示以下标 i 字符结尾的最长有效括号的长度
vector<int> dp(n, 0);
int maxLen = 0;
// 状态转移
for (int i = 1; i < n; i++) {
if (s[i] == ')') {
if (s[i - 1] == '(')
dp[i] = (i >= 2 ? dp[i - 2] : 0) + 2;
else if (i - dp[i - 1] > 0 && s[i - dp[i - 1] - 1] == '(')
dp[i] = dp[i - 1] + ((i - dp[i - 1] - 2) >= 0 ? dp[i - dp[i - 1] - 2] : 0) + 2;
}
maxLen = max(maxLen, dp[i]);
}
return maxLen;
}
};