Java：将 int[] 转换为范围的最小表示

Question

给定一组 int 值，如何将系列解析为计数序列符号？

例子：

{1, 2, 3, 4, 5, 9, 13, 14, 15} -> "1-5,9,13-15"
{4, 6, 8, 10, 11, 12, 15, 17}  -> "4,6,8,10-12,15,17"

我正在寻找一种能够产生这些结果的方法。这是我到目前为止所拥有的，但我现在非常难过：

测试代码：

import java.util.Arrays;
public class TestSequencing {
    public static void main(String[] args) {
        int[] numbers1 = {1, 2, 3, 4, 5, 9, 13, 14, 15};
        String numbers1s = "1-5,9,13-15";
        System.out.println(Arrays.toString(numbers1));
        System.out.println("Expected:\t" + numbers1s);
        System.out.println("Produced:\t" + sequenceNums(numbers1) + "\n");

        int[] numbers2 = {3, 5, 6, 9, 12};
        String numbers2s = "3,5-6,9,12";
        System.out.println(Arrays.toString(numbers2));
        System.out.println("Expected:\t" + numbers2s);
        System.out.println("Produced:\t" + sequenceNums(numbers2) + "\n");

        int[] numbers3 = {1, 2, 3, 4, 5, 6, 7};
        String numbers3s = "1-7";
        System.out.println(Arrays.toString(numbers3));
        System.out.println("Expected:\t" + numbers3s);
        System.out.println("Produced:\t" + sequenceNums(numbers3) + "\n");
    }

    public static String sequenceNums(int[] nums) {
        StringBuilder sb = new StringBuilder();
        int rangeStart = nums[0];
        int previous = nums[0];
        int current;
        int expected = previous + 1;

        for (int i = 1 ; i < nums.length ; i++) {
            current = nums[i];
            expected = previous + 1;               
            if (current != expected || i == (nums.length - 1)) {
                if (current == rangeStart) {
                    sb.append(previous + ",");
                } else {
                    sb.append(rangeStart + "-" + previous + ",");
                }                
                rangeStart = current;
            }              
            previous = current;
        }
        if (sb.charAt(sb.length() - 1) == ',') {
            sb.deleteCharAt(sb.length() - 1);
        }
        return sb.toString();
    }
}

输出：

[1, 2, 3, 4, 5, 9, 13, 14, 15]
Expected:   1-5,9,13-15
Produced:   1-5,9-9,13-14

[3, 5, 6, 9, 12]
Expected:   3,5-6,9,12
Produced:   3-3,5-6,9-9

[1, 2, 3, 4, 5, 6, 7]
Expected:   1-7
Produced:   1-6

Answer 1

试试这个：

private static void appendRange(StringBuilder sb, int begin, int end) {
    sb.append(",").append(begin);
    if (end != begin)
        sb.append("-").append(end);
}

public static String sequenceNums(int[] nums) {
    StringBuilder sb = new StringBuilder();
    if (nums.length == 0) return sb.toString();
    int begin = nums[0], end = nums[0];
    for (int cur : nums)
        if (cur - end <= 1)
            end = cur;
        else {
            appendRange(sb, begin, end);
            begin = end = cur;
        }
    appendRange(sb, begin, end);
    return sb.substring(1);
}

@Test
public void testSequenceNums() {
    assertEquals("1-5,9,13-15", sequenceNums(new int[] {1, 2, 3, 4, 5, 9, 13, 14, 15}));
    assertEquals("4,6,8,10-12,15,17", sequenceNums(new int[] {4, 6, 8, 10, 11, 12, 15, 17}));
    assertEquals("1-7", sequenceNums(new int[] {1, 2, 3, 4, 5, 6, 7}));
    assertEquals("", sequenceNums(new int[] {}));
}

Answer 2

在 for 循环中你有两个问题：

1) 第二个if 应该是if (previous == rangeStart) {

2) 你没有处理循环中的最后一个数字（i == (nums.length - 1)）。

我会使用以下代码来做到这一点：

public static String sequenceNums(int[] nums) {
    StringBuilder sb = new StringBuilder();

    int rangeStart = nums[0];
    int previous = nums[0];
    int current;
    int expected = previous + 1;
    int size = nums.length;

    for (int i = 1 ; i < size ; i++) {
        current = nums[i];
        expected = previous + 1;

        if (current != expected) {
            addRange(sb, rangeStart, previous);
            rangeStart = current;
        }

        previous = current;
    }
    addRange(sb, rangeStart, nums[size - 1]);

    return sb.toString();
}

private void addRange(StringBuilder sb, int from, int to) {
    if (sb.length() > 0) {
        sb.append(",");
    }
    if (from == to) {
        sb.append(from);
    } else {
        sb.append(from + "-" + to);
    }
}

Answer 3

这是您的固定代码。

public class TestSequencing {

    public static void main(String[] args) {
        int[] numbers1 = {1, 2, 3, 4, 5, 9, 13, 14, 15};
        String numbers1s = "1-5,9,13-15";
        System.out.println(Arrays.toString(numbers1));
        System.out.println("Expected:\t" + numbers1s);
        System.out.println("Produced:\t" + sequenceNums(numbers1) + "\n");

        int[] numbers2 = {3, 5, 6, 9, 12};
        String numbers2s = "3,5-6,9,12";
        System.out.println(Arrays.toString(numbers2));
        System.out.println("Expected:\t" + numbers2s);
        System.out.println("Produced:\t" + sequenceNums(numbers2) + "\n");

        int[] numbers3 = {1, 2, 3, 4, 5, 6, 7};
        String numbers3s = "1-7";
        System.out.println(Arrays.toString(numbers3));
        System.out.println("Expected:\t" + numbers3s);
        System.out.println("Produced:\t" + sequenceNums(numbers3) + "\n");
    }

    public static String sequenceNums(int[] nums) {
        StringBuilder sb = new StringBuilder();
        int rangeStart = nums[0];
        int previous = nums[0];
        int current;
        int expected = previous + 1;

        for (int i = 1 ; i < nums.length ; i++) {
            current = nums[i];
            expected = previous + 1;               
            if (current != expected || i == (nums.length - 1)) {
                if (current == rangeStart) {
                    sb.append(previous + ",");
                } else {
                    if(rangeStart != previous) {
                        if(i == nums.length - 1)
                            sb.append(rangeStart + "-" + current);
                        else
                            sb.append(rangeStart + "-" + previous + ",");
                    } else {
                        if(i == nums.length - 1)
                            sb.append(rangeStart + "," + current);
                        else
                            sb.append(rangeStart + ",");
                    }
                }                
                rangeStart = current;
            }              
            previous = current;
        }
        if (sb.charAt(sb.length() - 1) == ',') {
            sb.deleteCharAt(sb.length() - 1);
        }
        return sb.toString();
    }

}

问题是，如果当前值与范围起始值不同，则需要检查两种情况

i) 如果范围以相同的先前值开始。如果是这样，则不需要用范围分隔相同的数字（例如：9-9 没有意义。只有 9 才有意义）。另一种要处理的情况是到达数组末尾。如果到达数组末尾，即使它不在任何范围内，也应该在末尾添加它

ii) 否则，如果未到达数组末尾，则范围以先前值开始和结束。如果到达数组的末尾，那将是范围的结束值

Answer 4

我采用以下方法来表示范围内的整数数组。

注意：数字应预先按升序排序。

我们将有两个变量 start 和 current，我们将在迭代中使用它们来识别范围。 index 将是数组的当前索引。

一旦找到一个范围，我们将继续将其推送到 StringBuilder 中。

这是代码：

// We will take this set of integers 
int[] temp = new int[] { 0, 1, 4, 5, 8, 9, 11, 12, 13 };

// Helper variables
Integer start = null, current = null;

// The found range(s) will be stored in this variable.
StringBuilder rangeBuilder = new StringBuilder();

// The current index of the array in iteration
int index = 0;

do {
    // During the first iteration both start & current will be null. So setting the current index value to them.
    if (start == null) {
        start = current = temp[index];
    } else {
        // Checking if the index value is the next number of current.
        if (temp[index] == (current + 1)) {
            current = temp[index];
        } else {
            if (start == current) {
                rangeBuilder.append(start + ",");
            } else {
                rangeBuilder.append(start + "-" + current + ",");

            }
            start = current = temp[index];
        }

    }

    // Checking if we have reached the end of the array.
    if (index + 1 == temp.length) {
        if (start == current) {
            rangeBuilder.append(start);
        } else {
            rangeBuilder.append(start + "-" + current);
        }
    }

} while (index++ < temp.length - 1);

// Printing the range.
System.out.println("Range: " + rangeBuilder.toString());

说明：

我们以这个例子为例：

{ 0, 1, 4, 5, 8, 9, 11, 12, 13 }

注意：我们将代表 start -> s & current -> c

迭代 1：

条目： start = null，current = null，rangeBuilder = ""

sc
0,  1,  4,  5,  8,  9,  11,  12, 13
^

索引值是当前的下一个数字吗？没有

索引是否到达最后一个元素？没有

退出： start = 0, current = 0, rangeBuilder = ""

迭代 2：

条目： start = 0，current = 0，rangeBuilder = ""

s   c
0,  1,  4,  5,  8,  9,  11,  12, 13
    ^

索引值是当前的下一个数字吗？是的。所以我们将 c 从 0 -> 1 移动。

索引是否到达最后一个元素？没有

退出： start = 0，current = 1，rangeBuilder = ""

迭代 3：

条目： start = 0，current = 1，rangeBuilder = ""

        sc
0,  1,  4,  5,  8,  9,  11,  12, 13
        ^

索引值是当前的下一个数字吗？不，所以我们将 s 和 c 推入字符串生成器 (rangeBuilder)。在推送时，我们检查 s 和 c 是否相同以避免重复。然后我们将 s & c 移动到索引值。

索引是否到达最后一个元素？没有

退出： start = 4, current = 4, rangeBuilder = "0-1,"

迭代 4：

条目： start = 4，current = 4，rangeBuilder = "0-1,"

        s   c
0,  1,  4,  5,  8,  9,  11,  12, 13
            ^

索引值是当前的下一个数字吗？是的。所以我们将 c 从 4 -> 5 移动。

索引是否到达最后一个元素？没有

退出： start = 4, current = 5, rangeBuilder = "0-1,"

迭代 5：

条目： start = 4, current = 5, rangeBuilder = "0-1,"

                sc
0,  1,  4,  5,  8,  9,  11,  12, 13
                ^

索引值是当前的下一个数字吗？不，所以我们将 s 和 c 推入字符串生成器 (rangeBuilder)。在推送时，我们检查 s 和 c 是否相同以避免重复。然后我们将 s & c 移动到索引值。

索引是否到达最后一个元素？没有

退出： start = 8, current = 8, rangeBuilder = "0-1,4-5,"

迭代 6：

条目： start = 8, current = 8, rangeBuilder = "0-1,4-5,"

                s   c
0,  1,  4,  5,  8,  9,  11,  12, 13
                    ^

索引值是当前的下一个数字吗？是的。所以我们将 c 从 8 -> 9 移动。

索引是否到达最后一个元素？没有

退出： start = 8, current = 9, rangeBuilder = "0-1,4-5,"

迭代 7：

条目： start = 8，current = 9，rangeBuilder = "0-1,4-5,"

                        sc
0,  1,  4,  5,  8,  9,  11,  12, 13
                        ^

索引值是当前的下一个数字吗？不，所以我们将 s 和 c 推入字符串生成器 (rangeBuilder)。在推送时，我们检查 s 和 c 是否相同以避免重复。然后我们将 s & c 移动到索引值。

索引是否到达最后一个元素？没有

退出： start = 11, current = 11, rangeBuilder = "0-1,4-5,8-9,"

迭代 8：

条目： start = 11，current = 11，rangeBuilder = "0-1,4-5,8-9,"

                        s    c
0,  1,  4,  5,  8,  9,  11,  12, 13
                             ^

索引值是当前的下一个数字吗？是的。所以我们将 c 从 11 -> 12 移动。

索引是否到达最后一个元素？没有

退出： start = 11, current = 12, rangeBuilder = "0-1,4-5,8-9,"

迭代 9：

条目： start = 11，current = 12，rangeBuilder = "0-1,4-5,8-9,"

                        s        c
0,  1,  4,  5,  8,  9,  11,  12, 13
                                 ^

索引值是当前的下一个数字吗？是的。所以我们将 c 从 12 -> 13 移动。

索引是否到达最后一个元素？是的。所以我们将 s 和 c 推入 String Builder (rangeBuilder)。在推送时，我们检查 s 和 c 是否相同以避免重复。

迭代结束。字符串生成器 (rangerBuilder) 将具有以下值：0-1,4-5,8-9,11-13

请随时改进此代码:)

Answer 5

我能够通过引入布尔标志并重新设计测试连续数字的方法来解决您的问题。如果当前和下一个数字是连续的，则触发inRangeFlag 进行下一次迭代。请参阅下面的代码 cmets 以获得进一步的细分：

   public static String sequenceNums(int[] nums) {

    StringBuilder sb = new StringBuilder();
    int current;
    int next;
    boolean inRangeFlag = false;

    for (int i = 0; i < nums.length; i++) {

        current = nums[i];

        // TRUE: if element is not last element, because last number is
        // always appended.
        if (i < nums.length - 1) {

            next = nums[i + 1];

            // TRUE: if current element and next are consecutive
            if (next - current == 1) {

                // If rangeflag is false, the current number is the start
                // of a range. Append the number with hyphen.
                if (!inRangeFlag) {
                    sb.append(current + "-");
                }

                // Trigger inRange Flag for next iteration.
                inRangeFlag = true;

            } else {
                sb.append(current + ",");
                inRangeFlag = false; // Turn flag false because not inRange.
            }
        } else {
            sb.append(current);
        }
    }
    return sb.toString();
}