A report on the problem of sword-pointing offer (Java Edition)--Poker Shun Zi 44

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Analyze problems

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The key point of this topic is that the size of the king can be regarded as arbitrary numbers, then we think of this arbitrary number as 0 is also possible, because poker in 1-15 have numbers, so you can use 0 to represent the size of the king

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So we convert the problem to 5 numbers from 0-15 to see if the problem is continuous, because 0 can be considered as arbitrary numbers, so we can use 0 to fill the number of vacancies, such as 0,1,3 is also a continuous

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To remember the key point or this arbitrary number, consider a problem, that is, if the number of 5 in the number of 0 is greater than the number of vacancies, then the pit must be able to fill up, so must be continuous, if the number of 0 is less than the number of vacancies, then 0 the number is not enough to fill all the pits, so it is discontinuous

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This problem is converted to look at a sorted array of more than 0 or interval number of the problem, see, the problem is constantly in the conversion, and step by step cobwebs the final answer slowly surfaced

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Solve the problem

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First of all, we first sorted out the number of 5 , Arrays.sort (array), and then iterate through the array of 0 of the number of arrays, in addition, if you find two zeros, there is no need to continue, you can jump out of the loop

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for (int i=0;i<array.length;i++) {

if (array[i]==0) numof0++;

if (numof0==2) break;

}

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Alternatively, you can put the if judgment statement in the For loop, so that you can enter the number of times the loop body, reduce the number of traversal

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for (int i=0;i<array.length&& (array[i]==0) && (numof0!=2); i++) numof0++;

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Then start with the first index1 not 0 , int index1=numof0; compare index1 and index2=index1+1 gaps between adjacent two

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while (Index2<array.length) {

// If there are two of the same, then the instructions must not be continuous

if (Array[index1]==array[index2]) {

return false;

}

Numofdif+=array[index2]-array[index1]-1;

// implement two adjacent comparisons to move back

Index1=index2;

index2++;

}

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Finally, the number of 0 and the number of pits return (numof0>numofdif?true:false) are compared.

A report on the problem of sword-pointing offer (Java Edition)--Poker Shun Zi 44