[LeetCode] 163. Missing Ranges

You are given an inclusive range [lower, upper] and a sorted unique integer array nums, where all elements are in the inclusive range.

A number x is considered missing if x is in the range [lower, upper] and x is not in nums.

Return the smallest sorted list of ranges that cover every missing number exactly. That is, no element of nums is in any of the ranges, and each missing number is in one of the ranges.

Each range [a,b] in the list should be output as:

• "a->b" if a != b
• "a" if a == b

Example 1:

Input: nums = [0,1,3,50,75], lower = 0, upper = 99
Output: ["2","4->49","51->74","76->99"]
Explanation: The ranges are:
[2,2] --> "2"
[4,49] --> "4->49"
[51,74] --> "51->74"
[76,99] --> "76->99"

Example 2:

Input: nums = [], lower = 1, upper = 1
Output: ["1"]
Explanation: The only missing range is [1,1], which becomes "1".

Example 3:

Input: nums = [], lower = -3, upper = -1
Output: ["-3->-1"]
Explanation: The only missing range is [-3,-1], which becomes "-3->-1".

Example 4:

Input: nums = [-1], lower = -1, upper = -1
Output: []
Explanation: There are no missing ranges since there are no missing numbers.

Example 5:

Input: nums = [-1], lower = -2, upper = -1
Output: ["-2"]

Constraints:

• -109 <= lower <= upper <= 109
• 0 <= nums.length <= 100
• lower <= nums[i] <= upper
• All the values of nums are unique.

缺失的区间。

给你一个闭区间 [lower, upper] 和一个 按从小到大排序 的整数数组 nums ，其中元素的范围在闭区间 [lower, upper] 当中。

如果一个数字 x 在 [lower, upper] 区间内，并且 x 不在 nums 中，则认为 x 缺失。

返回 准确涵盖所有缺失数字 的 最小排序 区间列表。也就是说，nums 的任何元素都不在任何区间内，并且每个缺失的数字都在其中一个区间内。

这个题跟 228题 很像，可以一块做。这个题注意题目最后给出的数据范围，是会有溢出的情况的所以一开始就需要把 integer 改成 long 。遍历 nums 的数字，首先第一个数字 num 如果 == lower，那就 lower++，不断提高缺失的部分的下界，比如例子一中的 0 和 1；当 num 不等于 lower 的时候，此时又分两种情况，如果 num == lower + 1 说明只遗漏了一个数字，比如例子一中的 2，那么就把这一个数字加入结果集；还有一种情况是 num 很大，则需要返回一个 range，[lower, num - 1]。

时间O(n)

空间O(1)

Java实现

class Solution {
    public List<String> findMissingRanges(int[] nums, int lower, int upper) {
        List<String> res = new ArrayList<>();
        long alower = (long) lower;
        long aupper = (long) upper;
        for (int num : nums) {
            if (num == alower) {
                alower++;
            } else if (alower < num) {
                if (alower + 1 == num) {
                    res.add(String.valueOf(alower));
                } else {
                    res.add(alower + "->" + (num - 1));
                }
                alower = (long) num + 1;
            }
        }
        if (alower == aupper) {
            res.add(String.valueOf(alower));
        } else if (alower < aupper) {
            res.add(alower + "->" + aupper);
        }
        return res;
    }
}

 

[2023年] 官方更新了函数签名，思路还是一样。

时间O(n)

空间O(1)

Java实现

class Solution {
    public List<List<Integer>> findMissingRanges(int[] nums, int lower, int upper) {
        List<List<Integer>> res = new ArrayList<>();
        for (int num : nums) {
            if (num > lower) {
                res.add(Arrays.asList(lower, num - 1));
            }
            lower = num + 1;
        }
        if (lower <= upper) {
            res.add(Arrays.asList(lower, upper));
        }
        return res;
    }
}

 

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163. Missing Ranges

228. Summary Ranges

LeetCode 题目总结