1033. 移動石子直到連續
需要特判一下間隔爲1和間隔爲2的情況
class Solution {
public int[] numMovesStones(int a, int b, int c) {
int x1 = a, x2 = b, x3 = c;
a = Math.min(Math.min(x1,x2),x3);
c = Math.max(Math.max(x1,x2),x3);
b = x1+x2+x3-a-c;
int[] res = new int[2];
res[0] = Math.min(1,c-b-1) + Math.min(1,b-a-1);
if(b-a ==2 || c -b == 2) res[0] = Math.min(res[0],1);
res[1] = c-a-2;
return res;
}
}
1034. 邊框着色
注意題意，是聯通分量的邊界染色。直接dfs+判斷邊界即可。
class Solution {
int fx[] = {0,0,1,-1};
int fy[] = {1,-1,0,0};
int[][] vis;
int n,m;
boolean check(int x, int y)
{
if(x < 0 || y < 0 || x >= n || y >= m) return false;
return true;
}
void dfs(int [][] grid, int x, int y, int p)
{
for(int i = 0; i < 4; i++)
{
int ix = x + fx[i];
int iy = y + fy[i];
if(check(ix,iy) && grid[ix][iy] == p)
{
if(vis[ix][iy] == 0)
{
vis[ix][iy] = 1;
dfs(grid,ix,iy,p);
}
}
else {
vis[x][y] = -1;
}
}
}
public int[][] colorBorder(int[][] grid, int r0, int c0, int color) {
n = grid.length;
m = grid[0].length;
vis = new int[n][m];
for(int i = 0; i < n; i++)
for(int j = 0; j < m; j++)
vis[i][j] = 0;
int p = grid[r0][c0];
vis[r0][c0] = 1;
dfs(grid,r0,c0,p);
for(int i = 0; i < n; i++)
for(int j = 0; j < m; j++)
if(vis[i][j] == -1) grid[i][j] = color;
return grid;
}
}
1035. 不相交的線
直接DP, dp[i][j]表示第一行的前i個和第二行的前j個進行匹配的最優結果。dp方程見代碼！
class Solution {
public int maxUncrossedLines(int[] A, int[] B) {
int[][] dp = new int[A.length][B.length];
for(int i = 0; i < A.length; i++)
for(int j = 0; j < B.length; j++)
dp[i][j] = 0;
for(int i = 0; i < A.length; i++)
{
if(i != 0) dp[i][0] = dp[i-1][0];
if(A[i] == B[0]) dp[i][0] = Math.max(dp[i][0],1);
}
for(int i = 0; i < B.length; i++)
{
if(i != 0) dp[0][i] = dp[0][i-1];
if(A[0] == B[i]) dp[0][i] = Math.max(dp[0][i],1);
}
for(int i = 1; i < A.length; i++)
for(int j = 1; j < B.length; j++)
{
dp[i][j] = Math.max(dp[i][j],dp[i-1][j-1]);
dp[i][j] = Math.max(dp[i][j],dp[i-1][j]);
dp[i][j] = Math.max(dp[i][j],dp[i][j-1]);
if(A[i] == B[j])
dp[i][j] = Math.max(dp[i][j],dp[i-1][j-1]+1);
//System.out.println(i+" "+ j + " " + dp[i][j]);
}
return dp[A.length-1][B.length-1];
}
}
1036. 逃離大迷宮
這道題起初一看覺得是到簡單的搜索題，但是看到數據量就知道簡單的暴力搜索肯定會TLE，所以我們需要想想辦法，然後注意觀察題意，其中標註了障礙物最多200個，所以障礙物圍成的圈的直徑一定小於200，所以我們只需要考慮兩種情況，一種是障礙物將兩個點圍在了一起，那麼我們只需要簡單的dfs搜索就好了，限制一下搜索的直徑。第二種是障礙物只圍了一個點，所以我們需要分別從起點和終點出發，做兩個搜索，判斷其是否被圍住，即是否能到達直徑爲200的圈外，如果其都能到達圈外，則說明起點可以到達終點。
class Solution {
public int[] fx = {0,0,1,-1};
public int[] fy = {1,-1,0,0};
public int sx,sy;
public boolean[][] vis;
public boolean[][] g;
boolean check(int x, int y)
{
if(x < 0 || y < 0 || x >= 1e6 || y >= 1e6) return false;
return true;
}
boolean check1(int x,int y)
{
int ix = x-sx+200;
int iy = y-sy+200;
if(ix < 0 || iy < 0 || ix >= 400 || iy >= 400) return false;
return true;
}
boolean dfs1(int x, int y, int tx, int ty)
{
if(Math.abs(x-sx) + Math.abs(y-sx) >= 200) return false;
if(x == tx && y == ty) return true;
for(int i = 0; i < 4; i++)
{
int ix = x + fx[i];
int iy = y + fy[i];
if(check(ix,iy) && check1(ix,iy) && !vis[ix-sx+200][iy-sy+200] && !g[ix-sx+200][iy-sy+200])
{
vis[ix-sx+200][iy-sy+200] = true;
if(dfs1(ix,iy,tx,ty))
return true;
}
}
return false;
}
boolean dfs2(int x, int y)
{
if(Math.abs(x-sx) + Math.abs(y-sx) >= 200) return true;
for(int i = 0; i < 4; i++)
{
int ix = x + fx[i];
int iy = y + fy[i];
if(check(ix,iy) && check1(ix,iy) && !vis[ix-sx+200][iy-sy+200] && !g[ix-sx+200][iy-sy+200])
{
vis[ix-sx+200][iy-sy+200] = true;
if(dfs2(ix,iy))
return true;
}
}
return false;
}
public boolean isEscapePossible(int[][] blocked, int[] source, int[] target) {
boolean res;
vis = new boolean[400][400];
g = new boolean[400][400];
for(int i = 0; i < 400; i++)
for(int j = 0; j < 400; j++)
g[i][j] = vis[i][j] = false;
sx = source[0];sy = source[1];
for(int i = 0; i < blocked.length; i++)
{
int ix = blocked[i][0] - source[0] + 200;
int iy = blocked[i][1] - source[1] + 200;
if(check1(blocked[i][0],blocked[i][1])) g[ix][iy] = true;
}
vis[200][200] = true;
res = dfs1(source[0],source[1],target[0],target[1]);
if(res) return res;
res = true;
for(int i = 0; i < 400; i++)
for(int j = 0; j < 400; j++)
vis[i][j] = false;
vis[200][200] = true;
res = res && dfs2(source[0],source[1]);
sx = target[0]; sy = target[1];
for(int i = 0; i < 400; i++)
for(int j = 0; j < 400; j++)
g[i][j] = vis[i][j] = false;
for(int i = 0; i < blocked.length; i++)
{
int ix = blocked[i][0] - target[0] + 200;
int iy = blocked[i][1] - target[1] + 200;
if(check1(blocked[i][0],blocked[i][1])) g[ix][iy] = true;
}
res = res && dfs2(target[0],target[1]);
return res;
}
}