leetcode778. 水位上升的泳池中游泳
題目
解題的方法
dfs + 二分發
在這裏二分的意思是在取值的範圍類，進行二分的查找，看是否能夠到達 右下角的元素。如果能夠到達就ri（右邊的邊界） = mid縮小mid的值,如果不能夠，就擴大mid的範圍，就是le（左邊的邊界） = mid。當le == ri時 就是最小的值。
class Solution {
class Road{
int[][] board;
int X_len, Y_len;
int[] X = {1,-1,0,0};
int[] Y = {0,0,-1,1};
int res = 1000;
public Road(int l1, int l2) {
board = new int[l1][l2];
board[0][0] = 1;
X_len = l1;
Y_len = l2;
}
public boolean dfs(int x, int y, int[][] grid, int mid) {
int i = 0;
int a,b;
if(x == X_len - 1 && y == Y_len - 1) {
return true;
}
for(i = 0; i < 4; i++){
a = x + X[i];
b = y + Y[i];
if(a >= 0 && a < X_len && b < Y_len && b >= 0 && board[a][b] == 0 && grid[a][b] <= mid){
board[a][b] = 1;
if(dfs(a, b, grid, mid)){
return true;
}
}
}
return false;
}
public void clear(int l1, int l2){
board = new int[l1][l2];
}
}
public int swimInWater(int[][] grid) {
int le = Math.max(grid[0][0],grid[grid.length - 1][grid[0].length - 1]), ri = 50 * 50 - 1;
Road road = new Road(grid.length, grid[0].length);
while(le < ri){
int mid = (le + ri)>>1;
//System.out.println("le = "+le+" mid = "+mid+" ri = "+ri);
if(road.dfs(0, 0, grid, mid)) {
ri = mid;
}
else{
le = mid + 1;
}
road.clear(grid.length, grid[0].length);
}
return ri;
}
}
地雷
原博客
最小的值（val）開始進行遞增，如果值小於或等於val則把數組賦值爲val，然後繼續遍歷下去就只要右下角的值等於val就說明這個值是最小適合的值（因爲val是從最小的值進行遞增的）
class Solution {
int[] X = {0,0,-1,1};
int[] Y = {1,-1,0,0};
public void dfs(int m,int n, int m_len, int n_len, int val, int[][] grid){
if(grid[m][n] >= val){
return ;
}
grid[m][n] = val;
for(int i = 0; i < 4; i++){
int a = m + X[i];
int b = n + Y[i];
if(a >= 0 && a < m_len && b >= 0 && b < n_len){
dfs(a,b,m_len,n_len,val, grid);
}
}
}
public int swimInWater(int[][] grid) {
int i;
int m_len = grid.length, n_len = grid[0].length;
int val = Math.max(grid[0][0], grid[m_len - 1][n_len - 1]);
while(true){
dfs(0, 0, m_len, n_len, val + 1, grid);
if(grid[m_len - 1][n_len - 1] == val + 1){
return val;
}
val++;
}
}
}