977. 有序数组的平方

LeetCode题目

注意事项

• 无

class Solution {
public:
    vector<int> sortedSquares(vector<int>& nums) {
        int len = nums.size();
        vector<int> num1(len, 0);
        int left = 0;
        int right = nums.size() - 1;
        int pos = right;

        while(left < right) {
            int l2 = nums[left] * nums[left];
            int r2 = nums[right] * nums[right];
            if (l2 < r2) {
                num1[pos--] = r2;
                right--; 
            } 
            else if (l2 > r2) {
                num1[pos--] = l2;
                left++;
            }
            else {
                num1[pos--] = l2;
                left++;
                num1[pos--] = r2;
                right--;
            }
        }
        if (left == right) num1[pos] = nums[left] * nums[left];
        return num1;
    }
};

209.长度最小的子数组

LeetCode题目

注意事项

• 判断好双指针控制的滑动窗的变化
• 判断好边界条件/特殊情况

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

59. 螺旋矩阵 II

LeetCode题目

注意事项

• 想清楚矩阵流转的方向,以及转向时坐标变换的方式

class Solution {
public:
    vector<vector<int>> generateMatrix(int n) {
        vector<vector<int>> M(n, vector<int>(n, 0));
        int l = -1, r = n, u = 0, d = n;
        int x = 0, y = 0, num = n * n;
        int ind = 1;
        while(ind <= num) {
            while(y < r) 
                M[x][y++] = ind++;
            r--;
            y--;
            x++;
            while(x < d)
                M[x++][y] = ind++;
            d--;
            x--;
            y--;
            while(y > l)
                M[x][y--] = ind++;
            l++;
            y++;
            x--;
            while(x > u)
                M[x--][y] = ind++;
            u++;
            x++;
            y++;
        }
        return M;
    }
};