977.有序数组的平方

题目链接https://leetcode.cn/problems/squares-of-a-sorted-array/description/

• 题解：

/**
 * Note: The returned array must be malloced, assume caller calls free().
 */
int* sortedSquares(int* nums, int numsSize, int* returnSize) {
    //返回数组的大小和传入数组的大小相同
    *returnSize = numsSize;
    int* result = (int*)malloc(sizeof(int)*numsSize);
    int right = numsSize - 1;
    int left = 0;
    for(int index = numsSize - 1;index >= 0;index --){
        int leftResult = nums[left] * nums[left];
        int rightResult = nums[right] * nums[right];
        if(leftResult > rightResult){
            result[index] = leftResult;
            left ++;
        }else{
            result[index] = rightResult;
            right --;
        }
    }
    return result;
}

209.长度最小的子数组

题目链接https://leetcode.cn/problems/minimum-size-subarray-sum/description/

• 题解：

int minSubArrayLen(int target, int* nums, int numsSize) {
    //初始化长度最小的子数组长度为最大值，便于得到最小长度
    int result = INT_MAX;
    int sum = 0;
    //双指针法求最短数组长度
    int left = 0;
    int right = 0;
    //以右指针遍历数组尾部
    for(; right < numsSize ; right ++){
        sum += nums[right];
        //当sum值>=target，收缩右边界并保存当前长度，以便每次更新最短长度
        while(sum >= target){
            int temp = right - left + 1;
            result = result < temp ? result : temp;
            sum -= nums[left ++];
        }
    }
    return result == INT_MAX ? 0 : result;
}

59.螺旋矩阵II

题目链接https://leetcode.cn/problems/spiral-matrix-ii/description/

• 题解：

/**
 * Return an array of arrays of size *returnSize.
 * The sizes of the arrays are returned as *returnColumnSizes array.
 * Note: Both returned array and *columnSizes array must be malloced, assume caller calls free().
 */
int** generateMatrix(int n, int* returnSize, int** returnColumnSizes) {
    //tips:
    *returnSize = n;
    *returnColumnSizes = (int*)malloc(sizeof(int) * n);


    //初始化返回结果数组result,一级指针(行)
    int** result = (int**)malloc(sizeof(int*) * n);
    //初始化二级指针
    for(int i = 0; i < n; i ++){
        result[i] = (int*)malloc(sizeof(int) * n);
        //tips:
        (*returnColumnSizes)[i] = n;
    }


    //设置循环起始位置
    int startX = 0;
    int startY = 0;
    int offset = 1;
    int count = 1;
    //可能的奇数圈的中间的填充位置
    int middle = n / 2;
    //控制转大圈
    int loop = n / 2;
    while(loop){
        int i = startX;
        int j = startY;
        for(;j < n - offset; j++){
            result[i][j] = count++;
        }
        for(;i < n - offset; i++){
            result[i][j] = count++;
        }
        for(;j > startY; j--){
            result[i][j] = count++;
        }
        for(;i > startX; i--){
            result[i][j] = count++;
        }
        offset++;
        startX++;
        startY++;
        loop--;
    }
    //若为奇数，填充最后的元素
    if(n % 2){
        result[middle][middle] = count;
    }
    return result;
}