Algorithm--The maximum difference between two adjacent numbers

Maximum difference between two adjacent numbers Code Implementation

1  PackageCom.hzf.sort;2 3 Importorg.junit.Test;4 5 /**6 * There is a shape array A, please design an algorithm with the complexity of O (n), calculate the maximum difference between two adjacent numbers after sorting. 7  * 8 * Given the size n of an int array A and a, return the maximum difference. The array element is guaranteed to be more than 1. 9  * Ten * Test Sample: [1,2,5,4,6],5 One * return: 2 A  *  -  * @authorHzf -  * the  */ -  Public classGap { -      Public intMaxgap (int[] A,intN) { -         /** + * The first step is to find out the difference between the maximum and minimum values of this group of numbers - * Step Two: Open the right bucket according to the difference (the number of buckets is +1, which is guaranteed to have an empty bucket in the middle of the bucket) + * Step three: Put each number in the corresponding bucket, there will be empty barrels in the case A * Fourth Step: simply loop through the bucket, compare the difference between the smallest value in a bucket and the maximum in the previous bucket, and find the maximum difference at          */ -          -         //The first step is to find out the difference between the maximum and minimum values for this group of numbers -         intMin = a[0]; -         intmax = A[0]; -          for(inti=0; i<a.length; i++){ in             if(A[i] < min) min =A[i]; -             if(A[i] > max) max =A[i]; to         } +          -         //Step Two: Open the right bucket according to the difference (the number of buckets is +1, the number of buckets is greater than the length of the array, there are empty buckets in the middle) the         //The third step: Put each number in the corresponding bucket, there will be empty barrels in the case *         Boolean[] Hasnum =New Boolean[A.length+1];//determine if there is a number in the bucket $         int[] Maxnumbucket =New int[A.length+1];//record the maximum value of the current bucketPanax Notoginseng         int[] Minnumbucket =New int[A.length+1];//record the minimum value of the current bucket -          for(inti=0; i<a.length; i++){ the             intBucketnum = Bucket (A[i], a.length, Min, max);//in which bucket the current number should be placed +Maxnumbucket[bucketnum] = Hasnum[bucketnum]?Math.max (Maxnumbucket[bucketnum], a[i]): A[i]; AMinnumbucket[bucketnum] = Hasnum[bucketnum]?math.min (Minnumbucket[bucketnum], a[i]): A[i]; theHasnum[bucketnum] =true; +         } -          $         //The fourth step: simply loop through the bucket, compare the difference between the smallest value in a bucket and the maximum in the previous bucket, and find the maximum difference $         intresult = 0;//The final result -         intLeftbucketmax = maxnumbucket[0];//No. 0 Barrels must have a value . -         intRightbucketmin = 0; the         intStartbucket = 1;//start the traversal from bucket 1th . -         Wuyi          for(intI=startbucket; i<a.length+1; i++){ the             if(!hasnum[i])Continue; -Rightbucketmin =Minnumbucket[i]; Wu             if(Rightbucketmin-leftbucketmax >result) -result = Rightbucketmin-Leftbucketmax; AboutLeftbucketmax =Maxnumbucket[i]; $         } -          -         returnresult; -     } A      Public intBucketLongNumLongArrlength,LongMinLongMax) {//prevents multiplication of two int numbers, out of bounds +         /** the * 3,7,12,6,4 - * Minimum 3, maximum 12, each bucket range is (12-3)/5 $          *  the * Assuming the current value is 6, it should be placed in (int) ((6-3)/(9/5)) barrels the          */ the         return(int) ((Num-min)/((max-min) *1.0/arrlength));  the     } - @Test in      Public voidTest () { the         int[] arr =New int[]{3,7,19,6,4}; theSystem.out.println (Maxgap (arr,5)); About     } the}

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Algorithm--The maximum difference between two adjacent numbers