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How to further optimize after implementing insertion sort in java

Dec 29, 2020 am 10:12 AM
java insertion sort

How to further optimize after implementing insertion sort in java

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Normal insertion: operate from the second element of the array, when it is found When the previous element is larger than it, a swap operation is performed.

static int[] insertSort(int[] array){
        int len = array.length;
        for (int begin = 1; begin < len; begin++){
            int cur = begin;
            while (cur > 0 && array[cur] < array[cur-1]){
                int tmp = array[cur];
                array[cur] = array[cur-1];
                array[cur-1] = tmp;
                cur--;
            }
        }
        return array;
    }
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The first step of optimization: operate from the second element of the array. If it is found that the previous element is larger than it, move the previous element back until cur The pointed element is greater than or equal to its previous element. At this time, the position pointed by cur is the position where the element to be inserted should be inserted.

    static int[] insertSort2(int[] array){
        int len = array.length;
        for (int begin = 1; begin < len; begin++){
            int cur = begin;
            int tmp = array[cur];
            while (cur > 0 && array[cur] < array[cur-1]){
                array[cur] = array[cur-1];
                cur--;
            }
            array[cur] = tmp;
        }
        return array;
    }
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Second step optimization

The third algorithm is essentially the same as the second one. They all find the position to be inserted and move the elements, but the third algorithm reduces the number of comparisons through binary search, that is, the calls to the cmp function, and also reduces the calls to the swap function. It is faster to find the position where the current element should be inserted, and then move it, which improves efficiency.

static int[] insertSort3(int[] array){
        int len = array.length;

        for (int begin = 1; begin < len; begin++){
            int v = array[begin];
            int insertIndex = search(array,begin);
            // 将 [insertIndex, begin) 范围内的元素往右边挪动一个单位
            for (int i = begin; i > insertIndex; i--){
                array[i] = array[i-1];
            }
            array[insertIndex] = v;
        }
        return array;
    }
    static int search(int[] array, int index){
        int begin = 0;
        int end = index;
        while(begin < end){
            int mid = (begin+end) >> 1;
            if (array[index] < array[mid]){
                end = mid;
            }else{
                begin = mid+1;
            }
        }
        return begin;
    }
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It should be noted that after using binary search, only the number of comparisons is reduced, but the average time complexity of insertion sort is still O(n^2)

The effect of separating the moving operations:

    package com.mj.sort.cmp;import com.mj.sort.Sort;public class InsertionSort3<T extends Comparable<T>> extends Sort<T> {

	//	protected void sort() {//		for (int begin = 1; begin < array.length; begin++) {//			T v = array[begin];//			int insertIndex = search(begin);//			// 将 [insertIndex, begin) 范围内的元素往右边挪动一个单位			for (int i = begin - 1; i >= insertIndex; i--) {							}//			for (int i = begin; i > insertIndex; i--) {//				array[i] = array[i - 1];//			}//			array[insertIndex] = v;//		}//	}
	
	@Override
	protected void sort() {
		for (int begin = 1; begin < array.length; begin++) {
			insert(begin, search(begin));	//元素索引给你,你告诉这个位置的元素往哪插
		} 
	}
	
	/**
	 * 将source位置的元素插入到dest位置
	 * @param source
	 * @param dest
	 */
	private void insert(int source, int dest) {
		T v = array[source];
		for (int i = source; i > dest; i--) {
			array[i] = array[i - 1];
		}
		array[dest] = v;
	}
	
	/**
	 * 利用二分搜索找到 index 位置元素的待插入位置
	 * 已经排好序数组的区间范围是 [0, index)
	 * @param index
	 * @return
	 */
	private int search(int index) {
		int begin = 0;
		int end = index;
		while (begin < end) {
			int mid = (begin + end) >> 1;
			if (cmp(array[index], array[mid]) < 0) {
				end = mid;
			} else {
				begin = mid + 1;
			}
		}
		return begin;
	}}
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