leetcode -- 350、367

350.Intersection of Two Arrays II

Problem Description

Given two arrays, write a function to compute their intersection.

Example 1:

Input: nums1 = [1,2,2,1], nums2 = [2,2]
Output: [2,2]
Example 2:

Input: nums1 = [4,9,5], nums2 = [9,4,9,8,4]
Output: [4,9]
Note:

Each element in the result should appear as many times as it shows in both arrays.
The result can be in any order.

Solution Method

題目意思理解錯了，之前以爲交集是連續的。

void Partition(int *nums,int low, int high)
{
    if(high>low)
    {
        int pivotpos=QuickSort(nums,low,high);
        Partition(nums,pivotpos+1,high);
        Partition(nums,low,pivotpos-1);
    }
}
int QuickSort(int *nums,int low,int high)
{
    int pivot=nums[low];
    while(high>low)
    {
        while(high>low&&nums[high]>=pivot)high--;
        nums[low]=nums[high];
        while(high>low&&nums[low]<=pivot)low++;
        nums[high]=nums[low];
    }
    nums[low]=pivot;
    return high;
}
int* intersect(int* nums1, int nums1Size, int* nums2, int nums2Size, int* returnSize)
{
    int * resArr = (int *) malloc (sizeof (int) * (nums1Size + 1));
    Partition(nums1, 0, nums1Size - 1);
    Partition(nums2, 0, nums2Size - 1);
    
    int sub1 = 0, sub2 = 0, sub = 0;
    while (sub1 < nums1Size && sub2 < nums2Size)
    {
        if (nums1[sub1] == nums2[sub2])
        {
            resArr[sub ++] = nums1[sub1];
            sub1 ++;  sub2 ++;
        }
        else if (nums1[sub1] < nums2[sub2])
            sub1 ++;
        else if (nums1[sub1] > nums2[sub2])
            sub2 ++;
    }
    *returnSize = sub;
    return resArr;
}

367. Valid Perfect Square

Problem Description

Given a positive integer num, write a function which returns True if num is a perfect square else False.

Note: Do not use any built-in library function such as sqrt.

Example 1:

Input: 16
Output: true
Example 2:

Input: 14
Output: false

Solution Method

binary Search

bool isPerfectSquare(int num)
{
    long left = 1, right = num;
    if (num == 1)
        return true;
    while (left <= right)
    {
        long mid = (left + right) / 2;
        if (mid * mid == num)
            return true;
        else if (mid * mid > num)
            right = mid - 1;
        else if (mid * mid < num)
            left = mid + 1;
    }
    return false;
}