代码随想录刷题day02-矩阵

代码随想录刷题-矩阵

977. 有序数组的平方

左右指针的经典应用

左右下标元素平方相比，大的放新数组右边，比较简单，不赘述

class Solution {
public:
vector<int> sortedSquares(vector<int> &nums) {
	int len = nums.size();
    vector<int> res(len, 0);
    int j = len - 1;
    int left = 0;
    int right = len - 1;
    while (left <= right) {
        int leftValue = pow(nums[left], 2);
        int rightValue = pow(nums[right], 2);
        if (leftValue < rightValue) {
            res[j--] = rightValue;
            right--;
        } else {
            res[j--] = leftValue;
            left++;
        }
    }
    return res;
}
};

209. 长度最小的子数组

注意审题是大于等于目标值，不是等于目标值

滑动窗口的题

思路是先加上最右边的值，然后再循环判断是否大于等目标值，若大于等于则计算比较长度，然后减去左值让左下标往右移

最后返回

class Solution {
public:
int minSubArrayLen(int target, vector<int> &nums) {
    int left = 0;
    int right = 0;
    int min = INT_MAX;
    int sum = 0;
    while (right < nums.size())
    {
        sum += nums[right];
        while (sum >= target)
        {
            int len = right - left + 1;
            min = min < len ? min : len;
            sum -= nums[left++];
        }   
        right++;
    }
    return min == INT_MAX ? 0 : min;
}
};

59. 螺旋矩阵 II

这一题主要是针对边界问题的探讨

先定圈数，在每一圈探讨每条行、边的值，然后再探讨中间值，看代码很容易看懂

class Solution {
public:
vector<vector<int>> generateMatrix(int n) {
    vector<vector<int>> res(n, vector<int>(n, 0));
    int count = 1;
    for (int round = 1; round <= n / 2; round++) {
        for (int row = round - 1, column = round - 1; column < n - round; column++) {
            res[row][column] = count++;
        }
        for (int row = round - 1, column = n - round; row < n - round; row++) {
            res[row][column] = count++;
        }
        for (int row = n - round, column = n - round; column > round - 1; column--) {
            res[row][column] = count++;
        }
        for (int row = n - round, column = round - 1; row > round - 1; row--) {
            res[row][column] = count++;
        }
    }
    if (n % 2 == 1) {
        res[n / 2][n / 2] = count;
    }
    return res;
}
};