Title Description:
Enter an array of integers to implement a function that adjusts the order of the numbers in the array so that all the odd digits are placed in the first half of the array, all the even digits are located in the second half of the array, and the relative positions between the odd and odd, even and even, are guaranteed.
Ideas:
This problem first can think of a train of thought, using the idea of fast sorting, set the front and back two pointers, respectively, from the array before and after the end of the loop, the previous pointer encountered even, after the pointer encountered odd Exchange.
Public classSolution { Public voidReorderarray (int[] Array) { intleft = 0, right = array.length-1; while(Left <Right ) { while(Left < right &&!)IsEven (Array[left])) left++; while(Left < right &&IsEven (Array[right] ) Right++; Swap (array, left, right); } } Public BooleanIsEven (intele) { return(Ele & 1) = = 0; } Public voidSwapint[] arr,intIintj) {intt =Arr[i]; Arr[i]=Arr[j]; ARR[J]=T; }}
But the above solution is actually problematic, because fast sequencing is an unstable sort, and the subject requires stability. Then consider the bubble sort and insert sort. The bubble sort is nothing more than an odd case of swapping before the odd, the insertion sort is found to be an even number inserted in front of the first ...
Here is a solution based on the idea of insertion sorting:
Public classSolution { Public voidReorderarray (int[] Array) { if(Array = =NULL|| Array.Length <= 1) return ; for(inti=0;i<array.length;i++){ if(!IsEven (Array[i])) { inttemp=Array[i]; intJ=i-1; while(j>=0&&IsEven (Array[j])) {Array[j+1]=Array[j]; J--; } array[j+1]=temp; } } } Public BooleanIsEven (intele) { return(Ele & 1) = = 0; } }
Adjust the array order so that the odd digits are preceded by even numbers