228. Summary Ranges

Description

You are given a sorted unique integer array nums.

A range [a,b] is the set of all integers from a to b (inclusive).

Return the smallest sorted list of ranges that cover all the numbers in the array exactly. That is, each element of nums is covered by exactly one of the ranges, and there is no integer x such that x is in one of the ranges but not in nums.

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

• "a->b" if a != b
• "a" if a == b

Example 1:

Input: nums = [0,1,2,4,5,7]
Output: ["0->2","4->5","7"]
Explanation: The ranges are:
[0,2] --> "0->2"
[4,5] --> "4->5"
[7,7] --> "7"


Example 2:

Input: nums = [0,2,3,4,6,8,9]
Output: ["0","2->4","6","8->9"]
Explanation: The ranges are:
[0,0] --> "0"
[2,4] --> "2->4"
[6,6] --> "6"
[8,9] --> "8->9"


Constraints:

• 0 <= nums.length <= 20
• -231 <= nums[i] <= 231 - 1
• All the values of nums are unique.
• nums is sorted in ascending order.

Solutions

Solution 1: Two Pointers

We can use two pointers $i$ and $j$ to find the left and right endpoints of each interval.

Traverse the array, when $j + 1 < n$ and $nums[j + 1] = nums[j] + 1$, move $j$ to the right, otherwise the interval $[i, j]$ has been found, add it to the answer, then move $i$ to the position of $j + 1$, and continue to find the next interval.

Time complexity $O(n)$, where $n$ is the length of the array. Space complexity $O(1)$.

• class Solution {
public List<String> summaryRanges(int[] nums) {
List<String> ans = new ArrayList<>();
for (int i = 0, j, n = nums.length; i < n; i = j + 1) {
j = i;
while (j + 1 < n && nums[j + 1] == nums[j] + 1) {
++j;
}
}
return ans;
}

private String f(int[] nums, int i, int j) {
return i == j ? nums[i] + "" : String.format("%d->%d", nums[i], nums[j]);
}
}

• class Solution {
public:
vector<string> summaryRanges(vector<int>& nums) {
vector<string> ans;
auto f = [&](int i, int j) {
return i == j ? to_string(nums[i]) : to_string(nums[i]) + "->" + to_string(nums[j]);
};
for (int i = 0, j, n = nums.size(); i < n; i = j + 1) {
j = i;
while (j + 1 < n && nums[j + 1] == nums[j] + 1) {
++j;
}
ans.emplace_back(f(i, j));
}
return ans;
}
};

• class Solution:
def summaryRanges(self, nums: List[int]) -> List[str]:
def f(i: int, j: int) -> str:
return str(nums[i]) if i == j else f'{nums[i]}->{nums[j]}'

i = 0
n = len(nums)
ans = []
while i < n:
j = i
while j + 1 < n and nums[j + 1] == nums[j] + 1:
j += 1
ans.append(f(i, j))
i = j + 1
return ans


• func summaryRanges(nums []int) (ans []string) {
f := func(i, j int) string {
if i == j {
return strconv.Itoa(nums[i])
}
return strconv.Itoa(nums[i]) + "->" + strconv.Itoa(nums[j])
}
for i, j, n := 0, 0, len(nums); i < n; i = j + 1 {
j = i
for j+1 < n && nums[j+1] == nums[j]+1 {
j++
}
ans = append(ans, f(i, j))
}
return
}

• function summaryRanges(nums: number[]): string[] {
const f = (i: number, j: number): string => {
return i === j ? ${nums[i]} : ${nums[i]}->\${nums[j]};
};
const n = nums.length;
const ans: string[] = [];
for (let i = 0, j = 0; i < n; i = j + 1) {
j = i;
while (j + 1 < n && nums[j + 1] === nums[j] + 1) {
++j;
}
ans.push(f(i, j));
}
return ans;
}


• public class Solution {
public IList<string> SummaryRanges(int[] nums) {
var ans = new List<string>();
for (int i = 0, j = 0, n = nums.Length; i < n; i = j + 1) {
j = i;
while (j + 1 < n && nums[j + 1] == nums[j] + 1) {
++j;
}
}
return ans;
}

public string f(int[] nums, int i, int j) {
return i == j ? nums[i].ToString() : string.Format("{0}->{1}", nums[i], nums[j]);
}
}

• impl Solution {
pub fn summary_ranges(nums: Vec<i32>) -> Vec<String> {
if nums.is_empty() {
return vec![];
}

let mut ret = Vec::new();
let mut start = nums[0];
let mut prev = nums[0];
let mut current = 0;
let n = nums.len();

for i in 1..n {
current = nums[i];
if current != prev + 1 {
if start == prev {
ret.push(start.to_string());
} else {
ret.push(start.to_string() + "->" + &prev.to_string());
}
start = current;
prev = current;
} else {
prev = current;
}
}

if start == prev {
ret.push(start.to_string());
} else {
ret.push(start.to_string() + "->" + &prev.to_string());
}

ret
}
}