Array
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 $O(1)$.
Python3 Java C++ Go TypeScript Rust C#
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15 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
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17 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 ;
}
ans . add ( f ( nums , i , 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 ] );
}
}
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17 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 ;
}
};
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16 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
}
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15 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 ;
}
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37 impl Solution {
#[allow(dead_code)]
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
}
}
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17 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 ;
}
ans . Add ( f ( nums , i , 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 ]);
}
}
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