代码随想录day2 Java版_java八股代码随想录

1.有序数组的平方

因为数组本身有序，因此最大值只出现在两端，想到左右指针 

class Solution {
    public int[] sortedSquares(int[] nums) {
        int [] res = new int[nums.length];
        int i = 0, j = nums.length-1, k = nums.length-1;
        while (i <= j) {
            if (nums[i] * nums[i] >= nums[j] * nums[j]) {
                res[k--] = nums[i] * nums[i];
                i++;
            } else {
                res[k--] = nums[j] * nums[j];
                j--;
            }
        }
        return res;
    }
}

2.长度最小的子数组

使用滑动窗口思想优化，用双指针实现

class Solution {
    public int minSubArrayLen(int target, int[] nums) {
        int[] sum = new int[nums.length+1];
        for(int i = 1; i <= nums.length; i++) {
            sum[i] = sum[i-1] + nums[i-1];
        }
        int res = 1000000;
        int i = 0;
        for(int j = 1; j <= nums.length; j++) {
            
            while(sum[j]-sum[i] >= target && i <= j) {
                res = res < j-i?res:j-i ;
                i++;
            }
        }
        if (res == 1000000) return 0;
        return res;
    }
}

3.螺旋数组

就是纯模拟，按照题目描述顺时针遍历

class Solution {
    public int[][] generateMatrix(int n) {
        int[][] nums = new int[n][n];
        int offset = 1,count = 1, sx = 0, sy = 0;
        int loop = n / 2;
        while (loop-- > 0){
            int i = sy, j = sx;
            for (; j < n-offset; j++) {
                nums[sx][j] = count++;
            }
            for (; i < n-offset; i++) {
                nums[i][j] = count++;
            }
            for (; j > sx; j--) {
                nums[i][j] = count++;
            }
            for (; i > sy; i--) {
                nums[i][j] = count++;
            }
            offset++;
            sx++;
            sy++;
        }
        if (n % 2 == 1) {
            nums[n/2][n/2] = count;
        }
        return nums;
    }
}