08 2020 档案
摘要:link class Solution { public: #define LL long long const int mod=1E9+7; vector<vector<LL>> table; int numOfWays(vector<int>& nums) { int n=nums.size()
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摘要:link class Solution { public: int maxNonOverlapping(vector<int>& nums, int target) { unordered_map<int,int> pos; pos[0]=-1; int sum=0; int n=nums.size
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摘要:link class Solution { public: int longestAwesome(string s) { int n=s.size(); vector<int> dp(1<<10,n); int res=1; dp[0]=-1; int mask=0; for(int i=0;i<n
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摘要:link 随着i增加,缺失的个数非减,在i处缺失的个数为arr[i]-(i+1).二分找到第一个缺失个数大于等于k的位置left, 则left-1处缺失的个数<k。left-1处缺失的个数为arr[left-1]-left, 还差k-arr[left-1]+left个,则答案是k+left. cla
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摘要:link class Solution { public: struct TrieNode{ TrieNode* children[26]; int id; TrieNode(){ id=-1; for(int i=0;i<26;i++) children[i]=nullptr; } }; Trie
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摘要:link prove: @interviewrecipes https://leetcode.com/problems/minimum-swaps-to-arrange-a-binary-grid/discuss/768076/Min-Adjacent-Swaps-to-Sort-the-array
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