【LeetCode 热题 100】打家劫舍 / 零钱兑换 / 单词拆分 / 乘积最大子数组 / 最长有效括号

⭐️个人主页：@小羊 ⭐️所属专栏：LeetCode 热题 100 很荣幸您能阅读我的文章，诚请评论指点，欢迎欢迎 ~

---

爬楼梯

• 爬楼梯

class Solution {
public:int climbStairs(int n) {int a = 0, b = 0, c = 1;for (int i = 1; i <= n; i++){a = b;b = c;c = a + b;}return c;}
};

---

杨辉三角

• 杨辉三角

class Solution {
public:vector<vector<int>> generate(int numRows) {vector<vector<int>> res(numRows);for (int i = 0; i < numRows; i++){res[i].resize(i + 1, 1);for (int j = 1; j < i; j++){res[i][j] = res[i - 1][j - 1] + res[i - 1][j];}}return res;}
};

---

打家劫舍

• 打家劫舍

class Solution {
public:int rob(vector<int>& nums) {int n = nums.size();vector<int> f(n), g(n);f[0] = nums[0];for (int i = 1; i < n; i++){f[i] = g[i - 1] + nums[i];g[i] = max(g[i - 1], f[i - 1]);}return max(f[n - 1], g[n - 1]);}
};

---

完全平方数

• 完全平方数

这是一个完全背包问题。

class Solution {
public:int numSquares(int n) {int m = sqrt(n);vector<int> dp(n + 1, 0x3f3f3f3f);dp[0] = 0;for (int i = 1; i <= m; i++){for (int j = i * i; j <= n; j++){dp[j] = min(dp[j], dp[j - i * i] + 1);}}return dp[n];}
};

---

零钱兑换

• 零钱兑换

class Solution {
public:int coinChange(vector<int>& coins, int amount) {const int N = 0x3f3f3f3f;vector<int> dp(amount + 1, N);dp[0] = 0;for (int i = 0; i < coins.size(); i++){for (int j = coins[i]; j <= amount; j++){dp[j] = min(dp[j], dp[j - coins[i]] + 1);}}return dp[amount] >= N ? -1 : dp[amount];}
};

---

单词拆分

• 单词拆分

class Solution {
public:bool wordBreak(string s, vector<string>& wordDict) {unordered_set<string> hashset(wordDict.begin(), wordDict.end());int n = s.size();s = ' ' + s;vector<bool> dp(n + 1);dp[0] = true;for (int i = 1; i <= n; i++){for (int j = 1; j <= i; j++){if (dp[j - 1] && hashset.count(s.substr(j, i - j + 1))){dp[i] = true;break;}}}return dp[n];}
};

---

最长递增子序列

• 最长递增子序列

class Solution {
public:int lengthOfLIS(vector<int>& nums) {int n = nums.size();vector<int> dp(n, 1);int res = 1;for (int i = 1; i < n; i++){for (int j = 0; j < i; j++){if (nums[j] < nums[i])dp[i] = max(dp[i], dp[j] + 1);res = max(res, dp[i]);}}return res;}
};

贪心解法。

class Solution {
public:int lengthOfLIS(vector<int>& nums) {int n = nums.size();vector<int> v;for (auto e : nums){if (v.empty() || e > v.back()) v.push_back(e);else{int l = 0, r = v.size() - 1;while (l < r){int mid = (r + l) / 2;if (v[mid] < e) l = mid + 1;else r = mid;}v[l] = e;}}return v.size();}
};

---

最大子数组和

• 最大子数组和

dp[i] 表示以 i 位置为结尾的所有连续子数组的最大和。

class Solution {
public:int maxSubArray(vector<int>& nums) {int n = nums.size();vector<int> dp(n + 1);int res = -0x3f3f3f3f;for (int i = 1; i <= n; i++){dp[i] = max(dp[i - 1], 0) + nums[i - 1];res = max(res, dp[i]);}return res;}
};

class Solution {
public:int maxSubArray(vector<int>& nums) {int pre = 0, res = -0x3f3f3f3f;for (auto e : nums){pre = (pre, 0) + e;res = max(res, pre);}return res;}
};

---

乘积最大子数组

• 乘积最大子数组

class Solution {
public:int maxProduct(vector<int>& nums) {int n = nums.size(), res = -0x3f3f3f3f;vector<int> f(n + 1, 1), g(n + 1, 1);for (int i = 1; i <= n; i++){int x = nums[i - 1];int y = x * f[i - 1], z = x * g[i - 1];f[i] = max(x, max(y, z));g[i] = min(x, min(y, z));res = max(res, f[i]);}return res;}
};

---

分割等和子集

• 分割等和子集

class Solution {
public:bool canPartition(vector<int>& nums) {int sum = 0;for (auto e : nums) sum += e;if (sum % 2) return false;sum /= 2;vector<bool> dp(sum + 1);dp[0] = true;for (int i = 0; i < nums.size(); i++){for (int j = sum; j >= nums[i]; j--){dp[j] = dp[j] || dp[j - nums[i]];}}return dp[sum];}
};

---

最长有效括号

• 最长有效括号

初始化 -1 处理第一个字符就是 ) 的情况，避免栈操作错误。
栈中存储未匹配的（ 下标或无效 ）下标。

class Solution {
public:int longestValidParentheses(string s) {stack<int> st;st.push(-1);int len = 0;for (int i = 0; i < s.size(); i++){if (s[i] == '(') st.push(i);else{st.pop();if (st.empty()) st.push(i);else len = max(len, i - st.top());}}return len;}
};

---

本篇文章的分享就到这里了，如果您觉得在本文有所收获，还请留下您的三连支持哦~