Merge Two Sorted Arrays without additional memory

最新推荐文章于 2022-04-28 23:08:10 发布

原创最新推荐文章于 2022-04-28 23:08:10 发布 · 176 阅读

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本文介绍了一种不使用额外内存的方法来合并两个有序数组到较长的数组中，通过从数组末尾开始比较并放置元素的方式实现。示例展示了如何将长度为3的长数组{1,3,5,0,0,0}

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原创转载请注明出处：http://agilestyle.iteye.com/blog/2361133

Question

Merge Two Sorted Arrays into the larger array, given the large array has extra storage. Merge without additional memory

Requirements

Long sorted array {1,3,5,0,0,0} (0 means blank spaces), so used length is 3

The short sorted array {2,4}

The expected merged array will be {1,2,3,4,5,0}

Additionally, no additional memory is used

e.g. declaring a new array is not allowed

Solution

1. Loop from end to beginning.

2. Keep track of two tail index values of two merged arrays.

3. Final-merged-tail position is equal to longTail+ShortTail+1 (both index start from 0!!)

4. If the shortTail is larger than or equal to 0 after merging, adding the remaining values to the merged array

package org.fool.java.test;

public class MergeTwoSortedArrays {
    public static void main(String[] args) {
        int[] longArray = {1, 3, 5, 7, 0, 0, 0, 0};
        int used = 4;
        int[] shortArray = {2, 4, 6};

        for (int i : longArray) {
            System.out.print(i + " ");
        }

        System.out.println();

        merge(longArray, shortArray, used);

        for (int i : longArray) {
            System.out.print(i + " ");
        }
    }

    // we passed 3 arguments, the first 2 are the 2 arrays
    // the longUsed is to present how many items are used in the longArray
    // This method returns a int value representing how many valid items to be counted in the merged long array
    private static int merge(int[] longArray, int[] shortArray, int longUsed) {
        int longTail = longUsed - 1;
        int shortTail = shortArray.length - 1;

        // the order is to merge from end to beginning
        while (longTail >= 0 && shortTail >= 0) {
            // check which one in the two arrays should be put at this current tail position
            if (longArray[longTail] > shortArray[shortTail]) {
                // which means at this position, we'd better to put the value int the longArray for the merged array
                longArray[longTail + shortTail + 1] = longArray[longTail];

                // notice the key here for the final index in the merged array
                // longTail + shortTail + 1 will be the final index, thinking that both tail indexes start from 0

                longTail--; // after we put one value in longArray, we need to shift our focus to the left a little
            } else {
                longArray[longTail + shortTail + 1] = shortArray[shortTail];
                shortTail--;
            }
        }

        // case 1, longTail not equal to 0? No problem, they are in position already
        // case 2, shortTail not equal to 0? Need to add element one by one to the final array
        while (shortTail >= 0) {
            longArray[shortTail] = shortArray[shortTail];   // notice the final merging is same as copying
            shortTail--;    // do not forget updating the index
        }

        // return how many used elements in the new merged array
        return shortArray.length + longUsed;
    }
}

Console Output