岛屿类问题的广度优先深度优先双解法（Leetcode题解-Python语言）
695. 島嶼的最大面積
先上最經(jīng)典的題目，詳細(xì)思路看這題的官方題解，簡(jiǎn)單來(lái)說(shuō)的島嶼問(wèn)題就是遍歷二維數(shù)組，一般都是從一塊陸地開(kāi)始，進(jìn)行深度優(yōu)先或者廣度優(yōu)先搜索，每次上下左右四個(gè)方向選其一然后尋找下一塊陸地，最后找到一整個(gè)島嶼（一片陸地）。
廣度優(yōu)先搜索：
class Solution:def maxAreaOfIsland(self, grid: List[List[int]]) -> int:ans = 0nr = len(grid)if nr == 0:return 0nc = len(grid[0])for r in range(nr):for c in range(nc):if grid[r][c] == 1:queue = collections.deque([(r, c)]) # 隊(duì)列grid[r][c] = 2cur = 1while queue:row, col = queue.popleft() # 先進(jìn)先出for x, y in [(row - 1, col), (row + 1, col), (row, col - 1), (row, col + 1)]:if 0 <= x < nr and 0 <= y < nc and grid[x][y] == 1:cur += 1queue.append((x, y))grid[x][y] = 2ans = max(ans, cur)return ansnr是行的長(zhǎng)度，nc是列的長(zhǎng)度，遍歷整個(gè)二維數(shù)組，如果遇到陸地if grid[r][c] == 1就開(kāi)始找島嶼。廣度優(yōu)先搜索用的是隊(duì)列，記錄的是二維坐標(biāo)。grid[r][c] = 2的目的是防止重復(fù)遍歷。cur是當(dāng)前島嶼的面積。從一塊陸地出發(fā)，如果它的四個(gè)方向[(row - 1, col), (row + 1, col), (row, col - 1), (row, col + 1)]的下一個(gè)坐標(biāo)（x, y）在范圍內(nèi)并且為陸地的話(huà)，就把它加入隊(duì)列中，同時(shí)標(biāo)記已遍歷，cur也加1。ans記錄最大的島嶼面積。
深度優(yōu)先搜索：
class Solution:def maxAreaOfIsland(self, grid: List[List[int]]) -> int:ans = 0nr = len(grid)if nr == 0:return 0nc = len(grid[0])for r in range(nr):for c in range(nc):if grid[r][c] == 1:stack = [(r, c)] # 棧grid[r][c] = 2cur = 1while stack:row, col = stack.pop() # 先進(jìn)后出for x, y in [(row - 1, col), (row + 1, col), (row, col - 1), (row, col + 1)]:if 0 <= x < nr and 0 <= y < nc and grid[x][y] == 1:cur += 1stack.append((x, y))grid[x][y] = 2ans = max(ans, cur)return ans深度優(yōu)先搜索和廣度優(yōu)先搜索基本一樣，只是改用棧來(lái)實(shí)現(xiàn)，每次彈出的位置就是剛剛加入的，即一條路走到底，再往回走（深度）；廣度優(yōu)先則按順序彈出，一定把最初的位置遍歷完之后再遍歷新的，如同石頭落入水中產(chǎn)生漣漪（廣度）。
廣度優(yōu)先與深度優(yōu)先切換，只需要改一條初始化語(yǔ)句，一個(gè)pop函數(shù)，幾個(gè)變量名就行了。
200. 島嶼數(shù)量
廣度優(yōu)先搜索：
class Solution:def numIslands(self, grid: List[List[str]]) -> int:ans = 0nr = len(grid)if nr == 0:return 0nc = len(grid[0])for r in range(nr):for c in range(nc):if grid[r][c] == "1":queue = collections.deque([(r, c)])grid[r][c] = "2"ans += 1while queue:row, col = queue.popleft()for x, y in [(row - 1, col), (row + 1, col), (row, col - 1), (row, col + 1)]:if 0 <= x < nr and 0 <= y < nc and grid[x][y] == "1":queue.append((x, y))grid[x][y] = "2"return ans深度優(yōu)先搜索：
class Solution:def numIslands(self, grid: List[List[str]]) -> int:ans = 0nr = len(grid)if nr == 0:return 0nc = len(grid[0])for r in range(nr):for c in range(nc):if grid[r][c] == "1":stack = [(r, c)]grid[r][c] = "2"ans += 1while stack:row, col = stack.pop()for x, y in [(row - 1, col), (row + 1, col), (row, col - 1), (row, col + 1)]:if 0 <= x < nr and 0 <= y < nc and grid[x][y] == "1":stack.append((x, y))grid[x][y] = "2"return ans代碼框架基本相同，區(qū)別：這題的數(shù)字用字符來(lái)表示，是“1”而不是1；不需要記錄當(dāng)前島嶼的面積，只需要在進(jìn)入一塊新陸地找島嶼時(shí)ans加1即可，實(shí)際上比上一題簡(jiǎn)單。
694. 不同島嶼的數(shù)量
廣度優(yōu)先搜索：
class Solution:def numDistinctIslands(self, grid: List[List[int]]) -> int:ans = set()nr = len(grid)if nr == 0:return 0nc = len(grid[0])for r in range(nr):for c in range(nc):if grid[r][c] == 1:queue = collections.deque([(r, c)])grid[r][c] = 2path = 'a'while queue:row, col = queue.popleft()nxt = [(row - 1, col), (row + 1, col), (row, col - 1), (row, col + 1)]for i in range(4): # 為了記錄方向x, y = nxt[i]if 0 <= x < nr and 0 <= y < nc and grid[x][y] == 1:path += str(i)queue.append((x, y))grid[x][y] = 2path += 'a'ans.add(path)return len(ans)深度優(yōu)先搜索：
class Solution:def numDistinctIslands(self, grid: List[List[int]]) -> int:ans = set()nr = len(grid)if nr == 0:return 0nc = len(grid[0])for r in range(nr):for c in range(nc):if grid[r][c] == 1:stack = [(r, c)]grid[r][c] = 2path = 'a'while stack:row, col = stack.pop()nxt = [(row - 1, col), (row + 1, col), (row, col - 1), (row, col + 1)]for i in range(4): # 為了記錄方向x, y = nxt[i]if 0 <= x < nr and 0 <= y < nc and grid[x][y] == 1:path += str(i)stack.append((x, y))grid[x][y] = 2path += 'a'ans.add(path)return len(ans)這題找的是不同島嶼的數(shù)量，如何定義島嶼是否相同（模式）呢？題目說(shuō)明：兩個(gè)島嶼被認(rèn)為是相同的，當(dāng)且僅當(dāng)一個(gè)島嶼可以通過(guò)平移變換（不可以旋轉(zhuǎn)、翻轉(zhuǎn)）和另一個(gè)島嶼重合。由于我們找陸地時(shí)是從左上向右下找的，所以進(jìn)入某種形狀島嶼的位置是確定的，因此，可以用探索島嶼的路徑（方向組合）來(lái)表示這個(gè)島嶼（類(lèi)似動(dòng)作游戲的上下左右組合技能）。
具體方法就是用字符串記錄這個(gè)方向的組合，同時(shí)要避免不同形狀的島嶼出現(xiàn)相同路徑的情況，在每次遍歷時(shí)加入標(biāo)記path += 'a'，保證路徑的唯一性。
463. 島嶼的周長(zhǎng)
廣度優(yōu)先搜索：
class Solution:def islandPerimeter(self, grid: List[List[int]]) -> int:ans = 0nr = len(grid)if nr == 0:return 0nc = len(grid[0])for r in range(nr):for c in range(nc):if grid[r][c] == 1:stack = [(r, c)]grid[r][c] = 2while stack:row, col = stack.pop()for x, y in [(row - 1, col), (row + 1, col), (row, col - 1), (row, col + 1)]:if (x < 0 or x >= nr or y < 0 or y >= nc) or (0 <= x < nr and 0 <= y < nc and grid[x][y] == 0):ans += 1if 0 <= x < nr and 0 <= y < nc and grid[x][y] == 1:stack.append((x, y))grid[x][y] = 2return ans深度優(yōu)先搜索：
class Solution:def islandPerimeter(self, grid: List[List[int]]) -> int:ans = 0nr = len(grid)if nr == 0:return 0nc = len(grid[0])for r in range(nr):for c in range(nc):if grid[r][c] == 1:stack = [(r, c)]grid[r][c] = 2while stack:row, col = stack.pop()for x, y in [(row - 1, col), (row + 1, col), (row, col - 1), (row, col + 1)]:if (x < 0 or x >= nr or y < 0 or y >= nc) or (0 <= x < nr and 0 <= y < nc and grid[x][y] == 0):ans += 1if 0 <= x < nr and 0 <= y < nc and grid[x][y] == 1:stack.append((x, y))grid[x][y] = 2return ans本題中只有一個(gè)島嶼，注意周長(zhǎng)即島嶼的邊界，只會(huì)出現(xiàn)在陸地與大邊界或者海洋的交界處，即指向四個(gè)方向的下一個(gè)坐標(biāo)（x，y）有 if (x < 0 or x >= nr or y < 0 or y >= nc) or (0 <= x < nr and 0 <= y < nc and grid[x][y] == 0):越界或者是海洋。
827. 最大人工島
廣度優(yōu)先搜索：
class Solution:def largestIsland(self, grid: List[List[int]]) -> int:ans = 0nr = len(grid)if nr == 0:return 0nc = len(grid[0])area = {}index = 2for r in range(nr):for c in range(nc):if grid[r][c] == 1:queue = collections.deque([(r, c)]) # 隊(duì)列grid[r][c] = indexcur = 1while queue:row, col = queue.popleft() # 先進(jìn)先出for x, y in [(row - 1, col), (row + 1, col), (row, col - 1), (row, col + 1)]:if 0 <= x < nr and 0 <= y < nc and grid[x][y] == 1:cur += 1queue.append((x, y))grid[x][y] = indexarea[index] = curindex += 1ans = max(area.values() or [0])for r in range(nr):for c in range(nc):if grid[r][c] == 0:seen = set()for x, y in [(r - 1, c), (r + 1, c), (r, c - 1), (r, c + 1)]:if 0 <= x < nr and 0 <= y < nc and grid[x][y] > 1:seen.add(grid[x][y])ans = max(ans, 1 + sum(area[i] for i in seen))return ans深度優(yōu)先搜索：
class Solution:def largestIsland(self, grid: List[List[int]]) -> int:ans = 0nr = len(grid)if nr == 0:return 0nc = len(grid[0])area = {}index = 2for r in range(nr):for c in range(nc):if grid[r][c] == 1:stack = [(r, c)] # 隊(duì)列grid[r][c] = indexcur = 1while stack:row, col = stack.pop() # 先進(jìn)先出for x, y in [(row - 1, col), (row + 1, col), (row, col - 1), (row, col + 1)]:if 0 <= x < nr and 0 <= y < nc and grid[x][y] == 1:cur += 1stack.append((x, y))grid[x][y] = indexarea[index] = curindex += 1ans = max(area.values() or [0]) # 防止全1的情況for r in range(nr):for c in range(nc):if grid[r][c] == 0:seen = set()for x, y in [(r - 1, c), (r + 1, c), (r, c - 1), (r, c + 1)]:if 0 <= x < nr and 0 <= y < nc and grid[x][y] > 1:seen.add(grid[x][y])ans = max(ans, 1 + sum(area[i] for i in seen))return ans一道困難題，基本思路是將現(xiàn)有的島嶼遍歷一次，用字典標(biāo)記不同的島嶼和島嶼面積，然后再遍歷所有的海洋格子，填充該格子后的大島嶼面積就是這個(gè)格子四個(gè)方向的任兩個(gè)島嶼面積之和加1，找出最大值即可。
1034. 邊界著色
遞歸法：
class Solution:def colorBorder(self, grid: List[List[int]], row: int, col: int, color: int) -> List[List[int]]:m, n = len(grid), len(grid[0])explored = set()def dfs(r, c):# 當(dāng)前是遍歷過(guò)的連通分量if (r, c) in explored:return False# 當(dāng)前越界了，說(shuō)明上一次的在網(wǎng)格的邊界上if r < 0 or c < 0 or r == m or c == n:return True# 當(dāng)前不是連通分量，說(shuō)明上一次的與不在分量中的網(wǎng)格相鄰if grid[r][c] != grid[row][col]:return True# 當(dāng)前的是范圍內(nèi)的沒(méi)遍歷過(guò)的連通分量explored.add((r,c))isBorder = Falsefor dx, dy in (0,1),(1,0),(0,-1),(-1,0):if dfs(r+dx, c+dy):isBorder = Trueif isBorder:grid[r][c] = colorreturn Falsedfs(row, col)return grid迭代法：
class Solution:def colorBorder(self, grid: List[List[int]], row: int, col: int, color: int) -> List[List[int]]:m, n = len(grid), len(grid[0])explored = set()original_color = grid[row][col]stack = [(row, col)]while stack:r, c = stack.pop()explored.add((r,c))for dx, dy in (0,1),(1,0),(0,-1),(-1,0):# 下一步越界了，說(shuō)明(r,c)在網(wǎng)格的邊界上if r+dx < 0 or c+dy < 0 or r+dx == m or c+dy == n:grid[r][c] = color# 遍歷過(guò)的連通分量跳過(guò)elif (r+dx, c+dy) in explored:continue# 下一步不是連通分量，說(shuō)明(r,c)與不在分量中的網(wǎng)格相鄰elif grid[r+dx][c+dy] != original_color:grid[r][c] = colorelse:stack.append((r+dx, c+dy))return grid這題指定了遍歷的起點(diǎn) (row, col)，所以迭代法實(shí)現(xiàn)中就不能用 for 循環(huán)而要用 while 循環(huán)。從起點(diǎn)出發(fā)，記錄下遍歷過(guò)的節(jié)點(diǎn)，然后考察其四個(gè)方向的下一個(gè)節(jié)點(diǎn)：如果下一個(gè)的節(jié)點(diǎn)越界了，說(shuō)明當(dāng)前節(jié)點(diǎn)在邊界上，則需要著色；如果下一個(gè)節(jié)點(diǎn)是遍歷過(guò)的連通分量，則跳過(guò)；如果下一個(gè)節(jié)點(diǎn)不是連通分量，則需要著色；剩下的情況就是下一個(gè)節(jié)點(diǎn)是在邊界內(nèi)的、為遍歷過(guò)的連通分量，則加入棧。
1765. 地圖中的最高點(diǎn)
class Solution:def highestPeak(self, isWater: List[List[int]]) -> List[List[int]]:m, n = len(isWater), len(isWater[0])ans = [[water-1 for water in row] for row in isWater]q = collections.deque((i, j) for i, row in enumerate(isWater) for j, _ in enumerate(row) if ans[i][j] == 0)while q:i, j = q.popleft()for x, y in ((i+1, j), (i-1, j), (i, j+1), (i, j-1)):if 0 <= x < m and 0 <= y < n and ans[x][y] == -1:ans[x][y] = ans[i][j] + 1q.append((x, y))return ans廣度優(yōu)先遍歷，先把所有的 water 方格坐標(biāo)加入隊(duì)列作為起始點(diǎn)，然后遍歷它們上下左右的格子，如果是還沒(méi)被遍歷過(guò)的，就在 ans 中高度加 1 并且加入隊(duì)列，直到?jīng)]有格子為止。
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