LeetCode 209. 长度最小的子数组

子数组是一个连续的，很容易想到滑动窗口

class Solution:
    def minSubArrayLen(self, target: int, nums: List[int]) -> int:
        windowSum = 0
        left, right = 0, 0
        res = float('inf')
        while right < len(nums):
            push = nums[right]
            windowSum += push
            right += 1

            while windowSum >= target:
                res = min(res, right - left)
                pop = nums[left]
                windowSum -= pop
                left += 1
        return res if res != float('inf') else 0

拓展

如果 nums 数组中包含负数，则窗口扩大时元素和不见得就增大，窗口缩小时元素和不见得就减小，这种情况就不能单纯使用滑动窗口技巧了，可能需要混合动态规划和单调队列来做。

862. 和至少为 K 的最短子数组

LeetCode 1425. 带限制的子序列和

LeetCode 53. 最大子数组和

这里给出的是滑动窗口解法

class Solution:
    def maxSubArray(self, nums: List[int]) -> int:
        windowSum = 0
        res = float("-inf")
        left = right = 0
        while right < len(nums):
            push = nums[right]
            windowSum += push
            right += 1
            
            res = max(res, windowSum)

            while windowSum < 0:
                pop = nums[left]
                windowSum -= pop
                left += 1

        return res 

LeetCode 59. 螺旋矩阵 II

模拟题，做这一题之前可以先试试 LeetCode 54. 螺旋矩阵

class Solution:
    def generateMatrix(self, n: int) -> List[List[int]]:
        matrix = [[None for _ in range(n)] for _ in range(n)]
        idx = 1
        a, b, x, y = 0, 0, n - 1, n - 1
        for _ in range(n ** 2):
            for j in range(b, y + 1):
                matrix[a][j] = idx
                idx += 1
            a += 1
            for i in range(a, x + 1):
                matrix[i][y] = idx
                idx += 1
            y -= 1
            for j in range(y, b - 1, -1):
                matrix[x][j] = idx
                idx += 1
            x -= 1
            for i in range(x, a - 1, -1):
                matrix[i][b] = idx
                idx += 1
            b += 1
        return matrix

LeetCode 54. 螺旋矩阵

class Solution:
    def spiralOrder(self, matrix: List[List[int]]) -> List[int]:
        m, n = len(matrix) - 1, len(matrix[0]) - 1
        x, y = 0, 0
        res = []
        while x <= m and y <= n:
            for j in range(y, n + 1):
                if x <= m and y <= n:
                    res.append(matrix[x][j])
            x += 1
            for i in range(x, m + 1):
                if x <= m and y <= n:
                    res.append(matrix[i][n])
            n -= 1
            for j in range(n, y - 1, -1):
                if x <= m and y <= n:
                    res.append(matrix[m][j])
            m -= 1
            for i in range(m, x - 1, -1):
                if x <= m and y <= n:
                    res.append(matrix[i][y])
            y += 1
        return res          

KamaCoder 44. 开发商购买土地

前缀和思想

import sys

def find(matrix):
    m, n = len(matrix), len(matrix[0])
    preSum_matrix = [[None for _ in range(n)] for _ in range(m)]
    preSum_matrix[0][0] = matrix[0][0]
    for i in range(m):
        for j in range(n):
            up = preSum_matrix[i - 1][j] if i - 1 >= 0 else 0
            left = preSum_matrix[i][j - 1] if j - 1 >= 0 else 0
            left_up = preSum_matrix[i - 1][j - 1] if i - 1 >= 0 and j - 1 >= 0 else 0
            preSum_matrix[i][j] = matrix[i][j] + up + left - left_up
    res = float('inf')
    for i in range(m - 1):
        if res == 0:
            return 0
        res = min(res, abs(preSum_matrix[m - 1][n - 1] - preSum_matrix[i][n - 1] * 2))
    for j in range(n - 1):
        if res == 0:
            return 0
        res = min(res, abs(preSum_matrix[m - 1][n - 1] - preSum_matrix[m - 1][j] * 2))
    return res
        
if __name__ == "__main__":
    lines = sys.stdin.read().strip().splitlines()
    idx = 0
    res = []
    while idx < len(lines):
        m, n = map(int, lines[idx].strip().split())
        idx += 1
        matrix = []
        for _ in range(m):
            line = list(map(int, lines[idx].strip().split()))
            matrix.append(line)
            idx += 1
        res.append(str(find(matrix)))
    sys.stdout.write("\n".join(res))