4242
4343<p >注意:本题与主站 695  ; 题相同:  ; <a href =" https://leetcode-cn.com/problems/max-area-of-island/ " >https://leetcode-cn.com/problems/max-area-of-island/</a ></p >
4444
45- 4645## 解法
4746
4847<!-- 这里可写通用的实现逻辑 -->
@@ -105,7 +104,7 @@ def find(x):
105104
106105# 合并a和b所在的两个集合
107106p[find(a)] = find(b)
108- d[find(a)] = dinstance
107+ d[find(a)] = distance
109108```
110109
111110<!-- tabs:start -->
@@ -127,7 +126,7 @@ class Solution:
127126 for x, y in [[0 , 1 ], [0 , - 1 ], [1 , 0 ], [- 1 , 0 ]]:
128127 res += dfs(grid, i + x, j + y, m, n)
129128 return res
130-
129+ 131130 m, n = len (grid), len (grid[0 ])
132131 res = 0
133132 for i in range (m):
@@ -145,12 +144,12 @@ class Solution:
145144 m, n = len (grid), len (grid[0 ])
146145 p = list (range (m * n))
147146 size = [1 ] * (m * n)
148-
147+ 149148 def find (x ):
150149 if p[x] != x:
151150 p[x] = find(p[x])
152151 return p[x]
153-
152+ 154153 for i in range (m):
155154 for j in range (n):
156155 if grid[i][j] == 1 :
@@ -272,29 +271,36 @@ DFS:
272271
273272``` ts
274273function maxAreaOfIsland(grid : number [][]): number {
275- let m = grid .length , n = grid [ 0 ]. length ;
276- let res = 0 ;
277- for ( let i = 0 ;i < m ; ++ i ) {
278- for (let j = 0 ; j < n ; ++ j ) {
279- if ( grid [ i ][ j ] == 1 ) {
280- res = Math . max ( dfs ( grid , i , j ), res );
281- }
282- }
274+ let m = grid .length ,
275+ n = grid [ 0 ]. length ;
276+ let res = 0 ;
277+ for (let i = 0 ; i < m ; ++ i ) {
278+ for ( let j = 0 ; j < n ; ++ j ) {
279+ if ( grid [ i ][ j ] == 1 ) {
280+ res = Math . max ( dfs ( grid , i , j ), res );
281+ }
283282 }
284- return res ;
285- };
283+ }
284+ return res ;
285+ }
286286
287287function dfs(grid : number [][], i : number , j : number ): number {
288- let m = grid .length , n = grid [0 ].length ;
289- if (i < 0 || i > m - 1 || j < 0 || j > n - 1 || grid [i ][j ] == 0 ) {
290- return 0 ;
291- }
292- grid [i ][j ] = 0 ;
293- let res = 1 ;
294- for (let [dx, dy] of [[0 , 1 ], [0 , - 1 ], [1 , 0 ], [- 1 , 0 ]]) {
295- res += dfs (grid , i + dx , j + dy );
296- }
297- return res ;
288+ let m = grid .length ,
289+ n = grid [0 ].length ;
290+ if (i < 0 || i > m - 1 || j < 0 || j > n - 1 || grid [i ][j ] == 0 ) {
291+ return 0 ;
292+ }
293+ grid [i ][j ] = 0 ;
294+ let res = 1 ;
295+ for (let [dx, dy] of [
296+ [0 , 1 ],
297+ [0 , - 1 ],
298+ [1 , 0 ],
299+ [- 1 , 0 ],
300+ ]) {
301+ res += dfs (grid , i + dx , j + dy );
302+ }
303+ return res ;
298304}
299305```
300306
0 commit comments