Time: 20 minutes
First of all, it can be seen that this is a question to examine DFS
var swimInWater = function(grid) {
let row = grid.length; let col = grid[0].length;
var step = 0 while(true) {
for(let i = 0 ; i < row ; i ++) {
for(let j = 0 ; j < col ; j ++){
grid[i][j] -- }
}
var visited = []
for(let i = 0 ; i < row ; i ++) {
let arr = []
for(let j = 0 ; j < col ; j ++) {
arr.push(false)
}
visited.push(arr)
}
var dfs = function (i,j) {
if ( i< 0 || j < 0 || i >= row || j >= col) return false
if (visited[i][j]) {
return false } else {
visited[i][j] = true }
if (grid[i][j] > 0) return false if (i === row - 1 && j === col - 1 && grid[i][j] <= 0) return true return dfs(i+1,j) || dfs(i-1,j) || dfs(i,j-1)|| dfs(i,j + 1)
}
step++ if (dfs(0,0) === true) break
}
return step
};We don’t have to start from 1, we can use the binary search to find the leftmost insertion point.
var swimInWater = function(grid) {
let row = grid.length; let col = grid[0].length;
let right = -Infinity let left = Infinity for(let i = 0 ; i < row ; i ++) {
for(let j = 0 ; j < col ; j ++ ){
let num = grid[i][j]
right = Math.max(right,num)
left = Math.min(left,num)
}
}
let mid = left + Math.floor( (right - left) / 2 )
while(left <= right) {
mid = left + Math.floor( (right - left) / 2 )
var visited = []
var new_grid = []
for(let i = 0 ; i < row ; i ++) {
let arr = []
let _grid = []
for(let j = 0 ; j < col ; j ++) {
arr.push(false)
_grid.push(grid[i][j] - mid)
}
visited.push(arr)
new_grid.push(_grid)
}
var dfs = function (i,j) {
if ( i < 0 || j < 0 || i >= row || j >= col) return false
if (visited[i][j]) {
return false } else {
visited[i][j] = true }
if (new_grid[i][j] > 0) return false if (i === row - 1 && j === col - 1 && new_grid[i][j] <= 0) return true return dfs(i+1,j) || dfs(i-1,j) || dfs(i,j-1)|| dfs(i,j + 1)
}
if(dfs(0,0)) {
// Found the spare tire, continue to see if it can be smaller right = mid - 1 } else {
left = mid + 1 }
}
return left
};