LeetCode 热题 HOT 100 第二天

6. 盛水最多的容器

双指针

class Solution {
public:
    int maxArea(vector<int>& height) {
        int n = height.size();
        int l = 0, r = n - 1, res = 0;
        while (l < r) {
            res = max(res, min(height[l], height[r]) * (r - l));
            if (height[l] > height[r]) {
                r -- ;
            }
            else {
                l ++ ;
            }
        }
        return res;
    }
};

7. 三数之和

class Solution {
public:
    vector<vector<int>> threeSum(vector<int>& nums) {
        vector<vector<int>> res;
        sort(nums.begin(), nums.end());
        for (int i = 0; i < nums.size(); i ++ ) {
            if (i && nums[i] == nums[i - 1]) continue;
            for (int j = i + 1, k = nums.size() - 1; j < k; j ++ ) {
                if (j > i + 1 && nums[j] == nums[j - 1]) continue;
                while (j < k - 1 && nums[i] + nums[j] + nums[k - 1] >= 0) k -- ;
                if (nums[i] + nums[j] + nums[k] == 0) 
                    res.push_back({nums[i], nums[j], nums[k]});
            }
        }
        return res;
    }
};

8. 电话号码的字母组合

class Solution {
public:
    vector<string> res;
    string strs[10] = {
        "","","abc","def",
        "ghi","jkl","mno",
        "pqrs","tuv","wxyz"
    };
    vector<string> letterCombinations(string digits) {
        if (digits.empty()) return res;
        dfs(digits, 0, "");
        return res;
    }
    void dfs(string& digits, int u, string str) {
        if (u == digits.size()) res.push_back(str);
        else {
            for (auto c : strs[digits[u] - '0'])
                dfs(digits, u + 1, str + c);
        }
    }

};

9. 删除链表的倒数第n个结点

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* removeNthFromEnd(ListNode* head, int k) {
        auto dummy = new ListNode(-1);
        dummy -> next = head;
        auto l = dummy, r = head;
        for (int i = 0; i < k; i ++ ) r = r -> next;

        while (r) {
            r = r -> next;
            l = l -> next;
        }
        l->next = l->next->next;
        return dummy->next;
    }
};

10. 有效的括号

技巧

asc码差值$<=2$

class Solution {
public:
    bool isValid(string s) {
        stack<char> stk;

        for (auto& c : s) {
            if (c == '(' || c == '[' || c == '{') stk.push(c);
            else {
                if (stk.size() && abs(stk.top() - c) <= 2) stk.pop();
                else return false;
            }
        }

        return stk.empty();
    }
};