算法训练营 Day2 数组part02

一、209.长度最小的子数组

使用滑动窗口法，for循环里的变量是区间终点，for循环中用while逐步缩小区间，知道区间和小于target，终点继续向后移动

class Solution {
public:
    int minSubArrayLen(int target, vector<int>& nums) {
        int result = INT32_MAX;
        int i = 0;
        int sum = 0;
        int subLength = 0;
        for(int j=0; j < nums.size(); j++){
            sum += nums[j];
            while(sum >= target){
                subLength = j - i + 1;
                result = result < subLength ? result : subLength;
                sum -= nums[i++];
            }
        }
        return result==INT32_MAX ? 0 : result;
    }
};

二、59.螺旋矩阵II

四条边一圈是一个循环，循环n/2次，如果n是奇数，则中间需要额外填充

class Solution {
public:
    vector<vector<int>> generateMatrix(int n) {
        vector<vector<int>> res(n, vector<int>(n, 0));
        int startx = 0, starty = 0;
        int loop = n / 2;
        int mid = n / 2;
        int offset = 1;
        int count = 1;
        int i, j;
        while (loop--) {
            i = startx;
            j = starty;

            for (; j < n - offset; j++) {
                res[i][j] = count++;
            }
            for (; i < n - offset; i++) {
                res[i][j] = count++;
            }
            for (; j >= offset; j--) {
                res[i][j] = count++;
            }
            for (; i >= offset; i--) {
                res[i][j] = count++;
            }
            startx++;
            starty++;

            offset++;
        }
        if (n % 2 == 1) {
            res[mid][mid] = count;
        }
        return res;
    }
};