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Merged

pull merged 4 commits into AlgorithmAndLeetCode:master from youngyangyang04:master

Sep 9, 2022

Merged

[pull] master from youngyangyang04:master #83

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更新 0070.爬楼梯.md Java代码

fengxiuyang Sep 6, 2022

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

Pacifica-Git Sep 6, 2022

Merge pull request #1629 from zhicheng-lee/zhicheng-lee-patch-9

youngyangyang04 Sep 9, 2022

Merge pull request #1630 from Xin-Sheng-5075/master

youngyangyang04 Sep 9, 2022

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37 changes: 14 additions & 23 deletions problems/0070.爬楼梯.md

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Original file line number	Diff line number	Diff line change
Expand Up		@@ -213,22 +213,6 @@ public:


		### Java
	```Java
	class Solution {
	public int climbStairs(int n) {
	// 跟斐波那契数列一样
	if(n <= 2) return n;
	int a = 1, b = 2, sum = 0;

	for(int i = 3; i <= n; i++){
	sum = a + b;
	a = b;
	b = sum;
	}
	return b;
	}
	}
	```

		```java
		// 常规方式
Expand All		@@ -241,15 +225,22 @@ public int climbStairs(int n) {
		}
		return dp[n];
		}
	```

	```Java
		// 用变量记录代替数组
	public int climbStairs(int n) {
	int a = 0, b = 1, c = 0; // 默认需要1次
	for (int i = 1; i <= n; i++) {
	c = a + b; // f(i - 1) + f(n - 2)
	a = b; // 记录上一轮的值
	b = c; // 向后步进1个数
	class Solution {
	public int climbStairs(int n) {
	if(n <= 2) return n;
	int a = 1, b = 2, sum = 0;

	for(int i = 3; i <= n; i++){
	sum = a + b; // f(i - 1) + f(i - 2)
	a = b; // 记录f(i - 1),即下一轮的f(i - 2)
	b = sum; // 记录f(i),即下一轮的f(i - 1)
	}
	return b;
		}
	return c;
		}
		```

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33 changes: 33 additions & 0 deletions problems/0695.岛屿的最大面积.md

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Original file line number	Diff line number	Diff line change
Expand Up		@@ -167,3 +167,36 @@ public:
		};

		```

	# 其它语言版本

	## Java

	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。

	```java
	public int maxAreaOfIsland(int[][] grid) {
	int res = 0;
	for(int i = 0;i < grid.length;i++){
	for(int j = 0;j < grid[0].length;j++){
	//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
	if(grid[i][j] == 1){
	//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
	res = Math.max(res,dfs(grid,i,j));
	}
	}
	}
	return res;
	}
	public int dfs(int[][] grid,int i,int j){
	//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
	if(i < 0 \|\| i >= grid.length \|\| j < 0 \|\| j >= grid[0].length \|\| grid[i][j] == 0) return 0;
	//淹没土地,防止后续被重复计算
	grid[i][j] = 0;
	//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
	return 1 + dfs(grid,i - 1,j) +
	dfs(grid,i + 1,j) +
	dfs(grid,i,j + 1) +
	dfs(grid,i,j - 1);
	}
	```