代码随想录leetcode200题之数组

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1 介绍

本博客用来记录代码随想录leetcode200题中数组部分的题目。

2 训练

题目1：704二分查找

C++代码如下，

class Solution {
public:int search(vector<int>& nums, int target) {int res = -1;int l = 0, r = nums.size() - 1;//找到>=target的下标while (l <= r) {int mid = l + r >> 1;if (nums[mid] >= target) {res = mid;r = mid - 1;} else {l = mid + 1;}}if (res == -1) return res;else return nums[res] == target ? res : -1;}
};

python3代码如下，

class Solution:def search(self, nums: List[int], target: int) -> int:res = -1l, r = 0, len(nums) - 1while l <= r:mid = (l + r) // 2if nums[mid] >= target:res = mid r = mid - 1else:l = mid + 1if res == -1:return res else:return res if nums[res] == target else -1

题目2：27移除元素

C++代码如下，

class Solution {
public:int removeElement(vector<int>& nums, int val) {int i = 0;for (int j = 0; j < nums.size(); ++j) {if (nums[j] != val) nums[i++] = nums[j];}return i;}
};

python3代码如下，

class Solution:def removeElement(self, nums: List[int], val: int) -> int:i = 0for j in range(len(nums)):if nums[j] != val:nums[i] = nums[j]i += 1return i

题目3：977有序数组的平方

C++代码如下，

class Solution {
public:vector<int> sortedSquares(vector<int>& nums) {//从最大数开始排int n = nums.size();int i = 0, j = n - 1;vector<int> res(n);int k = n - 1;while (i <= j) {if (nums[i] * nums[i] > nums[j] * nums[j]) {res[k] = nums[i] * nums[i];i++;k--;} else {res[k] = nums[j] * nums[j];j--;k--;}}return res;}
};

python3代码如下，

class Solution:def sortedSquares(self, nums: List[int]) -> List[int]:n = len(nums)i, j = 0, n - 1k = n - 1res = [0] * n while i <= j:if nums[i] * nums[i] > nums[j] * nums[j]:res[k] = nums[i] * nums[i]i += 1k -= 1else:res[k] = nums[j] * nums[j]j -= 1k -= 1return res

题目4：209长度最小的子数组

C++代码如下，

//往大了讲，与子数组的最大和类似
class Solution {
public:int minSubArrayLen(int target, vector<int>& nums) {//滑动窗口法int res = INT_MAX;int s = 0;for (int r = 0, l = 0; r < nums.size(); ++r) {s += nums[r];while (s >= target) {res = min(res, r - l + 1);s -= nums[l++];}}return res == INT_MAX ? 0 : res;}
};

python3代码如下，

class Solution:def minSubArrayLen(self, target: int, nums: List[int]) -> int:s = 0l, r = 0, 0n = len(nums)res = 1e10while r < n:s += nums[r]while s >= target:res = min(res, r - l + 1)s -= nums[l]l += 1r += 1return res if res != 1e10 else 0

题目5：59螺旋矩阵 II

C++代码如下，

class Solution {
public:vector<vector<int>> generateMatrix(int n) {vector<vector<int>> res(n, vector<int>(n, 0));int val = 1;for (int k = 0; k <= n / 2; k += 1) {//起点(0 + k, 0 + k)int x0 = k, y0 = k;int length = n - 2 * k;//向右for (int step = 0; step < length - 1; ++step) {int x = x0;int y = y0 + step;res[x][y] = val++; }//向下x0 = x0;y0 = y0 + length - 1;for (int step = 0; step < length - 1; ++step) {int x = x0 + step;int y = y0;res[x][y] = val++;}//向左<--x0 = x0 + length - 1;y0 = y0;for (int step = 0; step < length - 1; ++step) {int x = x0;int y = y0 - step;res[x][y] = val++;}//向上x0 = x0;y0 = y0 - length + 1;for (int step = 0; step < length - 1; ++step) {int x = x0 - step;int y = y0;res[x][y] = val++;}//最后一步if (length == 1) {res[x0][y0] = val;}}return res;}
};

python3代码如下，

class Solution:def generateMatrix(self, n: int) -> List[List[int]]:res = [[0] * n for _ in range(n)]val = 1for k in range(n // 2 + 1):x0, y0 = k, klength = n - 2 * k#向右走-->for step in range(length-1):x, y = x0, y0 + step res[x][y] = val val += 1x0, y0 = x0, y0 + length - 1#向下走for step in range(length-1):x, y = x0 + step, y0 res[x][y] = val val += 1x0, y0 = x0 + length - 1, y0#向左走<--for step in range(length-1):x, y = x0, y0 - step res[x][y] = val val += 1x0, y0 = x0, y0 - length + 1#向上走for step in range(length-1):x, y = x0 - step, y0 res[x][y] = val val += 1if length == 1:res[x0][y0] = val val += 1return res 

3 参考

代码随想录官网

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