Leetcode题medium48/54/55/56/59/62/63/64,Python多种解法(四）

前文

继上一篇，[Leetcode题medium11/15/16/18/31/33/34/39/40,Python多种解法(三）(https://blog.csdn.net/weixin_42681866/article/details/88597930)，我们继续Array medium题目的解法分享。

48. Rotate Image


class Solution(object):"""利用zip的特性，可以针对集合里的元素进行第一个、第二个依次拼接，完美达到题目要求Runtime: 20 ms, faster than 100.00% of Python online submissions for Rotate Image.Memory Usage: 10.6 MB, less than 86.93% of Python online submissions for Rotate Image."""def rotate(self, matrix):""":type matrix: List[List[int]]:rtype: None Do not return anything, modify matrix in-place instead."""matrix[::] = zip(*matrix[::-1])class Solution1(object):"""先上下翻转，再以左上方到右下方斜线不动，进行翻转[[1,2,3],[4,5,6],[7,8,9]][[7,8,9],[4,5,6],[1,2,3]]Runtime: 20 ms, faster than 100.00% of Python online submissions for Rotate Image.Memory Usage: 11 MB, less than 5.30% of Python online submissions for Rotate Image."""def rotate(self, matrix):""":type matrix: List[List[int]]:rtype: void Do not return anything, modify matrix in-place instead."""if matrix:rows = len(matrix)cols = len(matrix[0])for i in range(rows // 2):for j in range(cols):matrix[i][j], matrix[rows - i - 1][j] = matrix[rows - i - 1][j], matrix[i][j]for i in range(rows):for j in range(i):matrix[i][j], matrix[j][i] = matrix[j][i], matrix[i][j]class Solution2(object):"""先调换4个角，再依次调换边Runtime: 24 ms, faster than 91.14% of Python online submissions for Rotate Image.Memory Usage: 10.8 MB, less than 23.67% of Python online submissions for Rotate Image."""def rotate(self, matrix):""":type matrix: List[List[int]]:rtype: void Do not return anything, modify matrix in-place instead."""n = len(matrix)for l in range(n // 2):r = n - 1 - lfor p in range(l, r):q = n - 1 - pcache = matrix[l][p]matrix[l][p] = matrix[q][l]matrix[q][l] = matrix[r][q]matrix[r][q] = matrix[p][r]matrix[p][r] = cache

54. Spiral Matrix


class Solution(object):"""讨论区大神一行解决，代码理解是and前后为真则返回后者，那当matrix为空的时候，执行pop会报错，此时返回matrixmatrix.pop(0)每次返回第一行数，而后段代码递归调用，以示例为例，第一次返回[(6, 9), (5, 8), (4, 7)]，因为被下一次pop出(6,9)，第二次返回[(8,7),(5,4)]，最后拼接即是答案Runtime: 40 ms, faster than 31.27% of Python3 online submissions for Spiral Matrix.Memory Usage: 13.2 MB, less than 5.18% of Python3 online submissions for Spiral Matrix."""def spiralOrder(self, matrix: List[List[int]]) -> List[int]:return matrix and [*matrix.pop(0)] + self.spiralOrder([*zip(*matrix)][::-1])class Solution2(object):"""正常思路，代码看的是真的舒服，即依次定义row、col，按照逻辑，先返回第一行，然后依次取每行的最后一个数，接着返回最后一行从后往前，最后返回每行的第一个数。然后循环此段操作。当l==r，说明只剩中间一行，直接reverse即可；当u==d，说明只剩中间一列，直接颠倒即可。Runtime: 36 ms, faster than 51.51% of Python3 online submissions for Spiral Matrix.Memory Usage: 13.2 MB, less than 5.18% of Python3 online submissions for Spiral Matrix."""def spiralOrder(self, matrix: List[List[int]]) -> List[int]:if not matrix or not matrix[0]:return []ans = []m, n = len(matrix), len(matrix[0])u, d, l, r = 0, m - 1, 0, n - 1while l < r and u < d:ans.extend([matrix[u][j] for j in range(l, r)])ans.extend([matrix[i][r] for i in range(u, d)])ans.extend([matrix[d][j] for j in range(r, l, -1)])ans.extend([matrix[i][l] for i in range(d, u, -1)])u, d, l, r = u + 1, d - 1, l + 1, r - 1if l == r:ans.extend([matrix[i][r] for i in range(u, d + 1)])elif u == d:ans.extend([matrix[u][j] for j in range(l, r + 1)])return ans

55. Jump Game


class Solution:"""贪心算法，每一步都算上一个数的索引和它的值的和是否有能力到达这一个数，如果可以的话，则说明有能力到达最后一个数，不行的话就直接返回FalseRuntime: 48 ms, faster than 73.92% of Python3 online submissions for Jump Game.Memory Usage: 14.7 MB, less than 5.28% of Python3 online submissions for Jump Game."""def canJump(self, nums: List[int]) -> bool:reach = 0for i in range(len(nums)):if i > reach:return Falsereach = max(reach, i+nums[i])return True

56. Merge Intervals


class Solution:"""题目的要求是针对itervals里的所有集合做区间判断，如果有重叠的区间就合并，那就有以下几种情况：1.[1,3],[2,4]==>第二个的start小于第一个的end，第一个的end小于第二个的end2.[1,4],[2,3]==>第二个的start小于第一个的end，第二个的end小于第一个的end3.[2,4],[1,3]==>第二个的start大于第一个的start，通过排序解决该情况，则转换为第1、2种4.[1,2],[3,4]==>即没有重叠空间所以如果排完序后，就剩1，2两种情况，本质是比较第一个和第二个的end，转为代码如下Runtime: 60 ms, faster than 80.12% of Python3 online submissions for Merge Intervals.Memory Usage: 15.9 MB, less than 5.34% of Python3 online submissions for Merge Intervals."""def merge(self, intervals: List[Interval]) -> List[Interval]:out = []for i in sorted(intervals, key=lambda i:i.start):if out and i.start <= out[-1].end:out[-1].end = max(i.end, out[-1].end)else:out.append(i)return out

59. Spiral Matrix II

class Solution:"""解法沿用54，Sprial Matrix的解法，定义u\d\l\r四个变量，以3*3为例，分为上、右、下、左四层，然后每层循环遍历出各自的数，最后再得出9所在的中心位置，如果超过3*3，则在while里循环一次Runtime: 40 ms, faster than 51.34% of Python3 online submissions for Spiral Matrix II.Memory Usage: 13.1 MB, less than 5.40% of Python3 online submissions for Spiral Matrix II."""def generateMatrix(self, n: int) -> List[List[int]]:matrix = [[0] * n for _ in range(n)]x, u, d, l, r = 1, 0, n - 1, 0, n - 1# x is the next value to write# u and d are upper and lower bound of current square/rectangle# l and r are left and right bound of current square/rectanglewhile l < r and u < d:for j in range(l, r):matrix[u][j] = xx += 1for i in range(u, d):matrix[i][r] = xx += 1for j in range(r, l, -1):matrix[d][j] = xx += 1for i in range(d, u, -1):matrix[i][l] = xx += 1u, d, l, r = u + 1, d - 1, l + 1, r - 1if l == r:matrix[u][r] = xreturn matrixclass Solution1(object):"""讨论区解法1,在上述解法上直接优化了三次遍历，牛皮！Runtime: 20 ms, faster than 100.00% of Python online submissions for Spiral Matrix II.Memory Usage: 10.8 MB, less than 26.97% of Python online submissions for Spiral Matrix II."""def generateMatrix(self, n):""":type n: int:rtype: List[List[int]]"""A = [[0] * n for _ in range(n)]i, j, di, dj = 0, 0, 0, 1for k in xrange(n*n):A[i][j] = k + 1if A[(i+di)%n][(j+dj)%n]:di, dj = dj, -dii += dij += djreturn Aclass Solution2(object):"""讨论区解法2，可以说很强大了Runtime: 20 ms, faster than 100.00% of Python online submissions for Spiral Matrix II.Memory Usage: 10.9 MB, less than 5.62% of Python online submissions for Spiral Matrix II."""def generateMatrix(self, n):""":type n: int:rtype: List[List[int]]"""A = [[n*n]]while A[0][0] > 1:A = [range(A[0][0] - len(A), A[0][0])] + zip(*A[::-1])return A

62. Unique Paths


class Solution(object):"""因为最后一步只有两种情况，要么从左到右，要么从上到下，所以可以去算前面两种情况的和解法事先定义一个集合，每一种路径则在目的地+1，到最后的[-1][-1]即能找到最后的路径总数Runtime: 16 ms, faster than 100.00% of Python online submissions for Unique Paths.Memory Usage: 10.8 MB, less than 31.15% of Python online submissions for Unique Paths."""def uniquePaths(self, m, n):""":type m: int:type n: int:rtype: int"""res = [[1 for _ in range(m)] for _ in range(n)]for i in range(1,n):for j in range(1,m):res[i][j] = res[i][j-1] + res[i-1][j]return res[-1][-1]class Solution(object):"""上面的优化！Runtime: 16 ms, faster than 100.00% of Python online submissions for Unique Paths.Memory Usage: 10.6 MB, less than 96.72% of Python online submissions for Unique Paths."""def uniquePaths(self, m, n):""":type m: int:type n: int:rtype: int"""dp = [1] + [0] * (n-1)for i in range(m):for j in range(1, n):dp[j] += dp[j-1]return dp[-1]

63. Unique Paths II

class Solution(object):"""直接按照62的解题思路来做Runtime: 24 ms, faster than 61.86% of Python online submissions for Unique Paths II.Memory Usage: 10.9 MB, less than 22.36% of Python online submissions for Unique Paths II."""def uniquePathsWithObstacles(self, obstacleGrid):""":type obstacleGrid: List[List[int]]:rtype: int"""if not obstacleGrid or not obstacleGrid[0]:return 0m = len(obstacleGrid)n = len(obstacleGrid[0])dp = [[0]*n for _ in range(m)]if obstacleGrid[0][0] == 0:dp[0][0] = 1else:return 0for i in range(m):for j in range(n):if obstacleGrid[i][j] == 1:continueif i > 0:dp[i][j] += dp[i-1][j]if j > 0:dp[i][j] += dp[i][j-1]return dp[-1][-1]

64. Minimum Path Sum


class Solution(object):"""题目容易和62、63弄混了，该题的意思是在给定的双重数组里，找出从gird[0][0]到grid[-1][-1]的最短距离所以原理还是和unique path相差不大，即最短距离就是当前的位置+左边/上边的最短路径，依次推导便是了即遍历两次，每次取左边或者上边的最小值，但要考虑两种特殊情况：第一行和第一列，因为他们各自没有上边和左边，因此当row=0，直接取左边的值为最小值，col=0.取右边的值为最小值Runtime: 32 ms, faster than 94.90% of Python online submissions for Minimum Path Sum.Memory Usage: 11.8 MB, less than 53.62% of Python online submissions for Minimum Path Sum."""def minPathSum(self, grid):""":type grid: List[List[int]]:rtype: int"""m,n = len(grid),len(grid[0])for row in range(m):for col in range(n):if row == col == 0:before = 0elif row == 0:before = grid[row][col-1]elif col == 0:before = grid[row-1][col]else:before = min(grid[row-1][col],grid[row][col-1])grid[row][col] = before + grid[row][col]return grid[-1][-1]

总结

本次分享到此结束，谢谢~

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Leetcode题medium48/54/55/56/59/62/63/64,Python多种解法(四）

前文

48. Rotate Image

54. Spiral Matrix

55. Jump Game

56. Merge Intervals

59. Spiral Matrix II

62. Unique Paths

63. Unique Paths II

64. Minimum Path Sum

总结

Leetcode题medium48/54/55/56/59/62/63/64,Python多种解法(四）相关推荐

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