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1. Yunqi Conference Limited-Time Rush Answer Match-Morning Field
The title of this session has the original title on LeetCode, and the link to the solution is as follows:
LeetCode 862. The shortest subarray whose sum is at least K (prefix and +deque monotonic stack)
2. Yunqi Conference Limited Time Rush Answer Match-Noon Field
Topic: Sliding Puzzle
description:
In a 3x3 grid, there are 8 boards numbered 1 to 8, and a space numbered 0.
One move can exchange space 0 with one of the four adjacent boards.
Given the initial and target board layout, return to the target layout with the least number of moves .
If you cannot move from the initial arrangement to the target arrangement, return -1.
样例1 ：
输入:
[
[2,8,3],
[1,0,4],
[7,6,5]
]
[
[1,2,3],
[8,0,4],
[7,6,5]
]
输出:
4
解释:
[ [
[2,8,3], [2,0,3],
[1,0,4], --> [1,8,4],
[7,6,5] [7,6,5]
] ]
[ [
[2,0,3], [0,2,3],
[1,8,4], --> [1,8,4],
[7,6,5] [7,6,5]
] ]
[ [
[0,2,3], [1,2,3],
[1,8,4], --> [0,8,4],
[7,6,5] [7,6,5]
] ]
[ [
[1,2,3], [1,2,3],
[0,8,4], --> [8,0,4],
[7,6,5] [7,6,5]
] ]
样例 2：
输入:
[[2,3,8],[7,0,5],[1,6,4]]
[[1,2,3],[8,0,4],[7,6,5]]
输出:
-1
With LeetCode 773. Slider Puzzle (BFS state transition map the shortest distance) is almost identical
Just change the code above slightly
- Breadth-first search, compress the 3x3 map into a string, record the status, and do not repeat visits
class Solution {
vector<vector<int>> dir = {
{
1,0},{
0,1},{
0,-1},{
-1,0}};
int m, n, i, j, k, step = 0, size, x, y;
public:
/**
* @param init_state: the initial state of chessboard
* @param final_state: the final state of chessboard
* @return: return an integer, denote the number of minimum moving
*/
int minMoveStep(vector<vector<int>> &board, vector<vector<int>> &final_state) {
m = board.size(), n = board[0].size();
string ans, state;
for(i = 0; i < 3; ++i)
for(j = 0; j < 3; ++j)
ans += '0'+final_state[i][j];
int x0, y0, xi, yi;
pair<int,int> xy0;//0的坐标
queue<string> q;
unordered_set<string> visited;//访问标记集合
state = boardToString(board);//初始状态
if(state == ans)
return step;
q.push(state);
visited.insert(state);
while(!q.empty())
{
step++;
size = q.size();
while(size--)
{
xy0 = stringToBoard(q.front(), board);//还原地图，并得到0的坐标
q.pop();
x0 = xy0.first;
y0 = xy0.second;
for(k = 0; k < 4; ++k)
{
//0可以4个方向交换
xi = x0+dir[k][0];
yi = y0+dir[k][1];
if(xi>=0 && xi<m && yi>=0 && yi<n)
{
swap(board[xi][yi], board[x0][y0]);//交换
state = boardToString(board);//新的状态
if(state == ans)
return step;
if(!visited.count(state))//没有出现过该地图
{
visited.insert(state);
q.push(state);
}
swap(board[xi][yi], board[x0][y0]);//还原现场
}
}
}
}
return -1;
}
string boardToString(vector<vector<int>>& board)
{
//地图转成字符串
string s;
for (i = 0; i < m; i++)
for(j = 0; j < n; j++)
s.push_back(board[i][j]+'0');
return s;
}
pair<int,int> stringToBoard(string &s, vector<vector<int>>& board)
{
//字符串还原成地图，并return 0的坐标，方便下次挪动
for (i = m-1; i >= 0; i--)
for(j = n-1; j >= 0; j--)
{
board[i][j] = s.back()-'0';
s.pop_back();
if(board[i][j] == 0)
x = i, y = j;
}
return make_pair(x, y);
}
};
1106ms C++
3. Yunqi Conference Limited-time Rush Answer-Evening Field
Topic: Map Jump
description:
For a given n×nmap, each cell has a height, and each time you can only move to the adjacent cell, and require these two cells 高度差不超过target.
You can't get out of the map.
Find the smallest target that satisfies from the upper left corner (0,0)to the lower right corner(n-1,n-1)
n < = 1000 < = a r r [ i ] [ j ] < = 100000 n <= 1000 <= arr[i][j] <= 100000 n<=1000<=arr[i][j]<=100000
例 1:
输入:[[1,5],[6,2]],
输出:4,
解释:
有2条路线:
1. 1 -> 5 -> 2 这条路线上target为4。
2. 1 -> 6 -> 2 这条路线上target为5。
所以结果为4
例 2:
输入:[[1,5,9],[3,4,7],[6,8,1]],
输出:6
Problem solving:
- Binary search
- DFS judges whether there is a path on which the absolute value of the difference between each adjacent point is <= target
class Solution {
vector<vector<int>> dir = {
{
1,0},{
0,1},{
-1,0},{
0,-1}};
int n;
public:
/**
* @param arr: the map
* @return: the smallest target that satisfies from the upper left corner (0, 0) to the lower right corner (n-1, n-1)
*/
int mapJump(vector<vector<int>> &arr) {
// Write your code here.
if(arr.empty()) return 0;
n = arr.size();
int l = 0, r = 100000, mid, ans;
bool way = false;
while(l <= r)
{
mid = (l + r)/2;
way = false;
vector<vector<bool>> vis(n, vector<bool>(n, false));
vis[0][0] = true;
ok(arr, 0, 0, vis, mid, way);//DFS
if(way)//有满足要求的路径
{
ans = mid;
r = mid-1;
}
else
l = mid+1;
}
return ans;
}
void ok(vector<vector<int>> &arr, int x, int y, vector<vector<bool>>& vis, int target, bool& way)
{
if(way) return;
if(x==n-1 && y==n-1)
{
way = true;
return;
}
int i, j, k;
for(k = 0; k < 4; ++k)
{
i = x + dir[k][0];
j = y + dir[k][1];
if(i >= 0 && i < n && j >= 0 && j < n && !vis[i][j] && abs(arr[i][j]-arr[x][y])<=target)
{
vis[i][j] = true;
ok(arr,i,j,vis,target,way);
}
}
}
};
101ms C++
Similar topics: LeetCode 1102. The path with the highest score (Priority Queue BFS/Minimax Binary Search)
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