Leetcode周赛从contest-121开始。(一般是10个contest写一篇文章)
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Contest 121 (题号981~984)(2019年1月27日)
链接:https://leetcode.com/contest/weekly-contest-121
总结:2019年2月22日补充的报告。当时不想写。rank:1093/3924,AC:2/4。还是太慢了。
【984】String Without AAA or BBB(第一题 4分)(Greedy, M)
给了两个数字,A 代表 A 个 ‘A’, B 代表 B 个‘B’ 在字符串里面。返回一个可行的字符串,字符串中包含 A 个‘A’, B 个 ‘B’,但是没有连续的 ‘AAA‘ 或者 ‘BBB‘。
题解:当时写的很长。现在说下greedy的方法,我们每次只想着只放一个a或者一个b,然后如果当前字符串的长度大于等于2,就需要判断前面是不是有两个连续的a或者连续的b。如果有的话,那么当前肯定不能放a或者放b。
1 class Solution { 2 public: 3 string strWithout3a3b(int A, int B) { 4 string res; 5 while (A > 0 || B > 0) { 6 if (res.size() >= 2) { 7 int size = res.size(); 8 if (res[size-1] == ‘a‘ && res[size-2] == ‘a‘) { 9 res += "b"; --B; 10 } else if (res[size-1] == ‘b‘ && res[size-2] == ‘b‘) { 11 res += "a"; --A; 12 } else if (A >= B) { 13 res += "a"; --A; 14 } else { 15 res += "b"; --B; 16 } 17 } else { 18 if (A >= B) { 19 res += "a"; --A; 20 } else { 21 res += "b"; --B; 22 } 23 } 24 } 25 return res; 26 } 27 };
【981】Time Based Key-Value Store(第二题 5分)
【983】Minimum Cost For Tickets(第三题 5分)(DP)
某个旅游景点有日票,周票和月票三种类型的票价。给了一个array,代表一年中的第几天去景点,问遍历完这个数组最少需要多少花费。
Train tickets are sold in 3 different ways:
- a 1-day pass is sold for
costs[0]
dollars; - a 7-day pass is sold for
costs[1]
dollars; - a 30-day pass is sold for
costs[2]
dollars.
Input: days = [1,4,6,7,8,20], costs = [2,7,15]
Output: 11
Explanation:
For example, here is one way to buy passes that lets you travel your travel plan:
On day 1, you bought a 1-day pass for costs[0] = $2, which covered day 1.
On day 3, you bought a 7-day pass for costs[1] = $7, which covered days 3, 4, ..., 9.
On day 20, you bought a 1-day pass for costs[0] = $2, which covered day 20.
In total you spent $11 and covered all the days of your travel.
题解:动态规划,dp[i] 代表前 i 天的最小花费。转移方程:
如果第 i 天不在访问的列表里面: dp[i] = dp[i-1]
else: dp[i] = min(dp[i-1] + cost[0], dp[i-7] + cost[1] + dp[i-30] + cost[2])
时间复杂度是: O(N)
1 class Solution { 2 public: 3 //dp[i] represents minimum money need cost in first i days 4 int mincostTickets(vector<int>& days, vector<int>& costs) { 5 const int totDays = 400; 6 vector<int> dp(totDays, INT_MAX); 7 dp[0] = 0; 8 set<int> st(days.begin(), days.end()); 9 for (int i = 1; i < totDays; ++i) { 10 if (st.find(i) == st.end()) { 11 dp[i] = dp[i-1]; 12 continue; 13 } 14 dp[i] = dp[i-1] + costs[0]; 15 dp[i] = min(dp[i], dp[max(i-7, 0)] + costs[1]); 16 dp[i] = min(dp[i], dp[max(i-30, 0)] + costs[2]); 17 } 18 return dp[366]; 19 } 20 };
【982】Triples with Bitwise AND Equal To Zero(第四题 7分)
Contest 122(题号985~988)(2019年2月3日,新春大礼包)
链接:https://leetcode.com/contest/weekly-contest-122
总结:这周题目非常简单,四题都过了。
【985】Sum of Even Numbers After Queries(第一题 4分)
给了一个数组,以及一个修改的序列。序列中的一个 pair 代表 p[0] 代表在原来数字上加上的值 val , p[1] 代表数组中需要修改元素的下标。求每次修改之后整个数组的偶数的和。
数据规模:
1 <= A.length <= 10000
-10000 <= A[i] <= 10000
1 <= queries.length <= 10000
-10000 <= queries[i][0] <= 10000
0 <= queries[i][1] < A.length
题解: 用一个变量curSum存储当前数组偶数的和,然后每次执行一个query之后,看当前元素是奇数变成偶数,还是偶数变成奇数,还是偶数变成偶数。来做不同的处理。
1 class Solution { 2 public: 3 vector<int> sumEvenAfterQueries(vector<int>& A, vector<vector<int>>& queries) { 4 const int n = A.size(), m = queries.size(); 5 vector<int> ret; 6 int curSum = 0; 7 for (auto num : A) { 8 if ((num & 1) == 0) { 9 curSum += num; 10 } 11 } 12 for (auto q : queries) { 13 int val = q[0], idx = q[1]; 14 int oriVal = A[idx]; 15 int newVal = oriVal + val; 16 if ((oriVal & 1) == 0) { 17 if (newVal & 1) { 18 curSum -= oriVal; 19 } else { 20 curSum += val; 21 } 22 } else { 23 if ((newVal & 1) == 0) { 24 curSum += newVal; 25 } 26 } 27 ret.push_back(curSum); 28 A[idx] = newVal; 29 } 30 return ret; 31 } 32 };
【988】Smallest String Starting From Leaf(第二题 5分)
给了一棵二叉树,树上的每个结点代表一个字母(0-25),问从叶子结点开始到根节点的字典序最小的单词是哪个?
题解:dfs 这颗树,把所有单词找出来。然后排序。注意这个题要求是从叶子结点开始,如果一个结点只有一个儿子,那么它本身不能算叶子结点,从它开始的单词要忽略掉。
1 /** 2 * Definition for a binary tree node. 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class Solution { 11 public: 12 string smallestFromLeaf(TreeNode* root) { 13 vector<string> words; 14 string str; 15 if (!root) { 16 return ""; 17 } 18 dfs(root, words, str); 19 sort(words.begin(), words.end()); 20 return words[0]; 21 } 22 void dfs(TreeNode* root, vector<string>& words, string& str) { 23 if (!root) { return; } 24 char c = root->val + ‘a‘; 25 str += string(1, c); 26 if (!root->left && !root->right) { 27 reverse(str.begin(), str.end()); 28 words.push_back(str); 29 reverse(str.begin(), str.end()); 30 str.pop_back(); 31 return; 32 } 33 if (root->left) { 34 dfs(root->left, words, str); 35 } 36 if (root->right) { 37 dfs(root->right, words, str); 38 } 39 str.pop_back(); 40 return; 41 } 42 };
【986】Interval List Intersections(第三题 5分)
给了两组左闭右闭的线段,返回这两组线段的交集。
Input: A = [[0,2],[5,10],[13,23],[24,25]], B = [[1,5],[8,12],[15,24],[25,26]]
Output: [[1,2],[5,5],[8,10],[15,23],[24,24],[25,25]]
Reminder: The inputs and the desired output are lists of Interval objects, and not arrays or lists.
题解:2 pointers 扫描一遍。
1 /** 2 * Definition for an interval. 3 * struct Interval { 4 * int start; 5 * int end; 6 * Interval() : start(0), end(0) {} 7 * Interval(int s, int e) : start(s), end(e) {} 8 * }; 9 */ 10 class Solution { 11 public: 12 vector<Interval> intervalIntersection(vector<Interval>& A, vector<Interval>& B) { 13 const int size1 = A.size(), size2 = B.size(); 14 int p1 = 0, p2 = 0; 15 vector<Interval> ret; 16 while (p1 < size1 && p2 < size2) { 17 Interval seg1 = A[p1], seg2 = B[p2]; 18 int s = max(seg1.start, seg2.start), e = min(seg1.end, seg2.end); 19 if (s > e) { 20 if (seg1.end > seg2.end) { p2++; } 21 else { p1++; } 22 continue; 23 } 24 Interval seg(s, e); 25 ret.push_back(seg); 26 if (seg1.end > seg2.end) { 27 p2++; 28 } else if (seg2.end > seg1.end) { 29 p1++; 30 } else { 31 p1++, p2++; 32 } 33 } 34 return ret; 35 } 36 };
【987】Vertical Order Traversal of a Binary Tree(第四题 5分)
给了一棵二叉树,给了树的结点上坐标的定义。根结点是 (0, 0), For each node at position (X, Y)
, its left and right children respectively will be at positions (X-1, Y-1)
and (X+1, Y-1)
. 按照 X 轴升序, Y 轴降序, 然后 value 升序的条件,返回这颗树的值。
题解:dfs一遍树,用map存起来。
1 /** 2 * Definition for a binary tree node. 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class Solution { 11 public: 12 vector<vector<int>> verticalTraversal(TreeNode* root) { 13 vector<vector<int>> ret; 14 if (!root) {return ret;} 15 dfs(root, 0, 0); 16 for (auto& pp : memo) { 17 int x = pp.first; map<int, vector<int>, greater<int>> mp = pp.second; 18 vector<int> temp; 19 for (auto& ele : mp) { 20 sort(ele.second.begin(), ele.second.end()); 21 for (auto& num : ele.second) { 22 temp.push_back(num); 23 } 24 } 25 ret.push_back(temp); 26 } 27 return ret; 28 } 29 map<int, map<int, vector<int>, greater<int>>> memo; // x-idx, list of values, and height 30 void dfs(TreeNode* root, int x, int y) { 31 if (!root) {return;} 32 memo[x][y].push_back(root->val); 33 dfs(root->left, x-1, y-1); 34 dfs(root->right, x+1, y-1); 35 } 36 37 };
Contest 123(题号985~988)(2019年2月10日)
Contest 124(题号993~996)(2019年2月17日)
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