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Signed-off-by: Xin-Sheng-5075 <2061461846@qq.com>

增加 695.岛屿的最大面积 Java 版本

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`‎problems/0695.岛屿的最大面积.md‎`

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`@@ -167,3 +167,36 @@ public:`
`167`	`167`	`};`
`168`	`168`
`169`	`169`	```
	`170`	`+`
	`171`	`+# 其它语言版本`
	`172`	`+`
	`173`	`+## Java`
	`174`	`+`
	`175`	`+这里使用深度优先搜索 DFS 来完成本道题目。我们使用 DFS 计算一个岛屿的面积,同时维护计算过的最大的岛屿面积。同时,为了避免对岛屿重复计算,我们在 DFS 的时候对岛屿进行 "淹没" 操作,即将岛屿所占的地方置为 0。`
	`176`	`+`
	`177`	+```java
	`178`	`+public int maxAreaOfIsland(int[][] grid) {`
	`179`	`+ int res = 0;`
	`180`	`+ for(int i = 0;i < grid.length;i++){`
	`181`	`+ for(int j = 0;j < grid[0].length;j++){`
	`182`	`+ //每遇到一个岛屿就计算这个岛屿的面积同时"淹没"这个岛屿`
	`183`	`+ if(grid[i][j] == 1){`
	`184`	`+ //每次计算一个岛屿的面积都要与res比较,维护最大的岛屿面积作为最后的答案`
	`185`	`+ res = Math.max(res,dfs(grid,i,j));`
	`186`	`+ }`
	`187`	`+ }`
	`188`	`+ }`
	`189`	`+ return res;`
	`190`	`+}`
	`191`	`+public int dfs(int[][] grid,int i,int j){`
	`192`	`+ //搜索边界:i,j超过grid的范围或者当前元素为0,即当前所在的地方已经是海洋`
	`193`	`+ if(i < 0 \|\| i >= grid.length \|\| j < 0 \|\| j >= grid[0].length \|\| grid[i][j] == 0) return 0;`
	`194`	`+ //淹没土地,防止后续被重复计算`
	`195`	`+ grid[i][j] = 0;`
	`196`	`+ //递归的思路:要求当前土地(i,j)所在的岛屿的面积,则等于1加上下左右相邻的土地的总面积`
	`197`	`+ return 1 + dfs(grid,i - 1,j) +`
	`198`	`+ dfs(grid,i + 1,j) +`
	`199`	`+ dfs(grid,i,j + 1) +`
	`200`	`+ dfs(grid,i,j - 1);`
	`201`	`+}`
	`202`	+```

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