Tech Life On The Way

Trapping Rain Water AC Rate: 165/540 My Submissions Given  n  non-negative integers representing an elevation map where the width of each bar is 1, compute how much water it is able to trap after raining. For example, Given  [0,1,0,2,1,0,1,3,2,1,2,1] , return  6 . Algorithm: scan from left to right get the leftMostHeight scan from right to left get the rightMost Height at the pos i, if(A[i] < min(left, right)) trap[i] = min(left, right) – A[i]; class Solution { public :        int trap( int A [], int n ) {               // Start typing your C/C++ solution below               // DO NOT write int main() function               if ( n < 3) return 0;        ...

Given a linked list, swap every two adjacent nodes and return its head. For example, Given   1->2->3->4 , you should return the list as   2->1->4->3 . Your algorithm should use only constant space. You may   not   modify the values in the list, only nodes itself can be changed. /**  * Definition for singly-linked list.  * struct ListNode {  *     int val;  *     ListNode *next;  *     ListNode(int x) : val(x), next(NULL) {}  * };  */ class Solution { public :     ListNode* swap(ListNode * head ) {         ListNode *tmp = head ->next;         tmp->next = head ;         head ->next = nullptr ;         return tmp;     }  ...

#include <iostream> using namespace std; #define N 4 void rotate90( int (* a )[N], int st , int ed ) {        if ( st >= ed ) return ;        int tmp;        for ( int i = 0 ; i < ed - st ; i++ ) { // index start from zero               // Note the rotate order               tmp = a [ st ][i+ st ];               a [ st ][i+ st ] = a [ ed -i][ st ];               a [ ed -i][ st ] = a [ ed ][ ed -i];               a [ ed ][ ed -i] = a [i+ st ][ ed ];               a [i+ st ][ ed ] = tmp;      ...

Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center). For example, this binary tree is symmetric:     1    / \   2   2  / \ / \ 3  4 4  3 But the following is not:     1    / \   2   2    \   \    3    3 /**  * Definition for binary tree  * struct TreeNode {  *     int val;  *     TreeNode *left;  *     TreeNode *right;  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}  * };  */ class Solution { public :        bool SymmetricUtil(TreeNode * n1 , TreeNode * n2 ) {               if ( n1 == nullptr && n2 == nullpt...

/* Given a sorted array, remove the duplicates in place such that each element appear only once and return the new length. Do not allocate extra space for another array, you must do this in place with constant memory. For example, Given input array A = [1,1,2], Your function should return length = 2, and A is now [1,2]. */ class Solution { public :        int findnext( int A [], int n , int st ) {               int id = st +1;               while (id < n ) {                      if ( A [ st ] != A [id]) break ;                      id++;            ...