[leetcode] sword finger offer 61. shunzi in playing cards

Draw at random from several sets of playing cards 5 card , Judge whether it's a down son , Is this 5 Are cards continuous .2～10 For the number itself ,A by 1,J by 11,Q by 12,K by 13, And big 、 Wang Wei 0 , It can be seen as any number .A Can't be regarded as 14.

Example 1:

Input : [1,2,3,4,5]
Output : True

Example 2:

Input : [0,0,1,2,5]
Output : True

Answer 1 ：

	/** *  The finger of the sword  Offer 61.  Shunzi in playing cards  */
    public boolean isStraight(int[] nums) {
    
        /* * 5  Cards should form shunzi , There can only be two duplicates at most  0, The remaining figures cannot be repeated , Return directly when repeating  false *  because  0  Can be used as any number of cards , So discuss the situation to clarify the thinking ： * 1.  Zero  0： The maximum number of points when forming a shunzi  -  Minimum points  === 4 * 2.  One  0： Such as  0,5,7,8,9, because  0  Can be viewed as  6, Then the maximum number of points （9） -  Minimum points （5） = 4 * 3.  Two  0： Such as  0,0,6,7,8, because  0  Can be viewed as  4,5  or  9,10, At this time, the maximum number of points （8） -  Minimum points （6） = 2 *  summary ： When the maximum number of cards  -  Minimum points  <= 4  Time can form shunzi  */
        Set<Integer> set = new HashSet<>();
        int max = 0;
        int min = 14;
        for (int num : nums) {
    
            //  If it is  0  Then continue to traverse , Find the maximum number of points and the minimum number of points 
            if (0 == num) {
    
                continue;
            }
            //  Determine whether the current element is repeated , Return directly if repeated  false
            if (!set.add(num)) {
    
                return false;
            }
            max = Math.max(max, num);
            min = Math.min(min, num);
        }
        return max - min <= 4;
    }

Explanation 2 ：

	/** *  The finger of the sword  Offer 61.  Shunzi in playing cards  */
    public boolean isStraight(int[] nums) {
    
        /* *  Sort the array first , Count the number of size kings in the process of traversing the array , *  And judge the Division  0  Whether there are duplicate points outside , Return directly if it exists  false *  notes ： After sorting the array , The maximum number of points must be  nums[4], The minimum number of points must be  nums[ Number of king and King ] * 1.  When there is no king of size, the minimum number of points must be  nums[0], Such as  1,2,3,4,5 * 2.  When there is a king of size, the minimum number of points must be  nums[1], Such as  0,1,2,4,5（ because  0  Can act as any point , Then it must act as a number larger than the minimum number of points in the current array ） * 3.  The minimum number of points must be  nums[2], Such as  0,0,2,4,5（ because  0  Can act as any point , Then it must act as a number larger than the minimum number of points in the current array ） */
        Arrays.sort(nums);
        int joker = 0;
        for (int i = 0; i < 4; i++) {
    
            //  Count the number of kings and kings 
            if (nums[i] == 0) {
    
                joker++;
            }
            //  There are duplicate points 
            else if (nums[i + 1] == nums[i]) {
    
                return false;
            }
        }
        //  The difference between the maximum number of points and the minimum number of points is less than or equal to  4  And shunzi 
        return nums[4] - nums[joker] <= 4;
    }

Limit ：

• The array length is 5
• The value of the array is [0, 13] .

source ： Power button （LeetCode）
link ：https://leetcode.cn/problems/bu-ke-pai-zhong-de-shun-zi-lcof