Title description
LL was in a very good mood today, because he bought a deck of playing cards and found that there were 2 big kings and 2 little kings in it (the deck of cards was originally 54 _ )... He randomly drew 5 cards from it and wanted to test it. Test his luck and see if he can draw a straight. If he can draw, he decides to buy a sports lottery, hehe! ! "Heart Ace, Spades 3, Little King, Big King, Square Piece 5", "Oh My God!" Not a straight... LL is not happy, he thought about it, and decided that Big \ Little King can be regarded as any number, and A is regarded as 1, J is 11, Q is 12, and K is 13. The 5 cards above can be turned into "1,2,3,4,5" (big and small kings are regarded as 2 and 4 respectively), "So Lucky!". LL decided to buy a sports lottery. Now, you are asked to use this card to simulate the above process, and then tell us how lucky LL is. If the card can form a straight, it will output true, otherwise it will output false. For convenience, you can consider the size king to be 0.
Code
import java.util.Arrays;
public class Solution {
public boolean isContinuous(int [] numbers) {
Arrays.sort(numbers);
int zorecount = 0;//0的个数
int sum =0;//非零元素之间差的和
int[] narr = new int[14];
if(numbers.length == 0){
return false;
}
//计算0的个数
for (int i = 0; i < numbers.length; i++) {
if (numbers[i] == 0){
zorecount++;
}
}
//判断非0重复元素情况
for (int i = 0; i < numbers.length; i++) {
narr[numbers[i]]++;
if (numbers[i] !=0 && narr[numbers[i]] > 1) {
return false;
}
}
//判断无大小王的情况
for (int i = 1; i < numbers.length; i++) {
if (numbers[0] != 0 && numbers[i] - numbers[i-1] != 1) {
return false;
}
}
//判断有大小王的情况
if (numbers[0] == 0) {
//计算非零元素之间差的和
for (int i = numbers.length-1; i > 0; i--) {
if(numbers[i] != 0 && numbers[i] - numbers[i-1] != numbers[i]){
sum += numbers[i] - numbers[i-1] -1;
}
}
if (zorecount < sum ) {
return false;
}
}
return true;
}
}
