# Problem 163: Missing Ranges

> <https://leetcode.com/problems/missing-ranges/>

![](/files/-Lpv9yfO_PhiOw8sbYPJ)

## 思路

* 这道题的麻烦之处就在于，是否有 lower -> 第一个元素， 最后一个元素 -> upper，以及什么时候是单个元素，什么时候要加箭头
* 我们可以用双指针来实现，prev 和 cur。当他们之间相差大于 1 的时候就证明有一段丢失了。

```java
public class Solution {
    public List<String> findMissingRanges(int[] nums, int lower, int upper) {
        List<String> rst = new ArrayList<String>();
        long prev = lower - 1;
        long cur = 0;
        for (int i = 0; i <= nums.length; i++) {
            cur = i == nums.length ? upper + 1 : nums[i];
            if (cur - prev > 1) {
                rst.add(getRange(prev + 1, cur - 1));
            }
            prev = cur;
        }

        return rst;
    }

    private String getRange(long start, long end) {
        return start == end ? String.valueOf(start) : start + "->" + end;
    }
}
```

## 易错点

1. 元素越界，所以用 long

   ```java
   [2147483647]
   0, 2147483647
   ```

   这个 test case 就会越界
2. 循环次数

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

   之所以是 `<=`是因为要考虑最后一个元素到 upper 这一段


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