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[转载]:合并两个已排序好的int数组,并排序返回c#实现

原作者: [db:作者] 来自: [db:来源] 收藏 邀请

/// <summary>
        /// 两个从小到大排序好的int数组,合并后也返回一个从小到大排序好的数组,
        /// 包含两个数组中全部的元素
        /// </summary>
        /// <param name="a"></param>
        /// <param name="b"></param>
        /// <returns></returns>
        public static int[] MergeArray(int[] a, int[] b)
        {
            if( a == null || b== null )
                throw new NotSupportedException();

            int lena = a.Length;
            int lenb = b.Length;
            int[] c = new int[lena+lenb];

            int i, j, n;
            i = j = n = 0;

            while (i < lena && j < lenb)
            {
                if (a[i] < b[j])
                {
                    c[n++] = a[i++];
                }
                else if (a[i] > b[j])
                {
                    c[n++] = b[j++];
                 }
                else
                {
                    c[n++] = a[i++];
                    c[n++] = b[j++];                   
                }
            }

            if (i == lena)
            {
                while (j < lenb)
                    c[n++] = b[j++];
            }
            else
            {
                while (i < lena)
                    c[n++] = a[i++];
            }

            return c;
        }

 

/// <summary>
        /// 两个从小到大排序好的int数组,合并后也返回一个从小到大排序好的数组,
        /// 包含两个数组中全部的元素
        /// </summary>
        /// <param name="a"></param>
        /// <param name="b"></param>
        /// <returns></returns>
        public static int[] MergeArray(int[] a, int[] b)
        {
            if( a == null || b== null )
                throw new NotSupportedException();

            int lena = a.Length;
            int lenb = b.Length;
            int[] c = new int[lena+lenb];

            int i, j, n;
            i = j = n = 0;

            while (i < lena && j < lenb)
            {
                if (a[i] < b[j])
                {
                    c[n++] = a[i++];
                }
                else if (a[i] > b[j])
                {
                    c[n++] = b[j++];
                 }
                else
                {
                    c[n++] = a[i++];
                    c[n++] = b[j++];                   
                }
            }

            if (i == lena)
            {
                while (j < lenb)
                    c[n++] = b[j++];
            }
            else
            {
                while (i < lena)
                    c[n++] = a[i++];
            }

            return c;
        }


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