题解是用了two pointer #include <cstdio> #include <cstdlib> #include <cstring> #include <cmath> #include <algorithm> #define IOS ios::sync_with_stdio(false) using namespace std; #define inf (0x3f3f3f3f) typedef long long int LL; #include <iostream> #include <sstream> #include <vect...
THIS ALGORITHM IS VERY USEFUL.IF WE SEE TYPES OF PROBLEM LIKE FIND OF 2 NUMBERS FROM ARRAY WHOSE SUM IS EQUAL TO X.IF WE DON'T KNOW WHAT IS TWO POINTER ALGORITHM THEN THIS PROBLEM CAN BE SOLVED BY O(n^2) (ONE FOR LOOP INSIDE THE ANOTHER)where n is no. of element in array by ...
32 #include<cstdio> #include<cstring> #include<algorithm> intmus[51000];//后缀最大值 inta[51000]; intb[51000]; intmain() { intn,k; scanf("%d%d",&n,&k); for(inti=1;i<=n;++i) scanf("%d",&b[i]); std::sort(b+1,b+1+n); intt1=n; for(inti=n;i;--i) { while(b...
题目分析:维护一个不小于m的数的个数的后缀和数组,可以枚举序列起点,二分查找右端点序列最近的一个<k,m>序列。因为最近右端点是不减的,所以也可以用two-pointer在O(n)的时间复杂度内得到结果。 代码如下: 使用二分查找: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 ...
twopointer排序后从头到尾扫一遍就行了。 AC代码: #include<cstdio> #include<iostream> #include<algorithm> #include<queue> #include #include<vector> #include<set> #include<string> #include<cmath> #include<cstring> #define ll long long #define...
}/*output sorted temporary array*/intk=0;inty =0;intnum=0;ofstream(argv[2]);//Pointer for inner vectorStringVector::iterator it2;for(StringVector2D::iterator outer = twoD.begin(); outer != twoD.end(); ++outer){ y++; k=0;for(it2= outer->begin(); it2!=outer->end(); ++i...
A simple algorithm if you use the linked list as a stack structure: #include <stdio.h> #include <stdlib.h> typedef struct list { int key; char value; struct list* next; } list; void print(list*); void add(list**, int, char); void reverse(list**); void deleteList(list*); int...
NeetCode makes no mention of this, but he obtains the correct algorithm for reasons that are unrelated to the intuition he provides. There are also other parts of the video where I disagree with his intuition (e.g., why $l=1$ and $r=n$ initially, which pointer to change when $a_l...
🎭 PsuendoCode Union Find Algorithm Pattern ♾ ⏰: O(V * logV) 🪐: O function find(int[] parent, i) { if (parent[i] == -1) return i; return find(parent, parent[i]); function union(int[] parent, x, y) { xset = find(parent, x); yset = find(parent, y); parent...
If we think similarly as we solved in the two sum problem, what we need to sort the arrays. We started from the beginning of one array and the end of another array. That was our two-pointer algorithm where we traversed the pointers based on the sum of the current...