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Update Merge Intervals solution and difficulty #287
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302b4da
Update deprecated solution with new method signature and removal of I…
nhantrivinh1 ad81028
Mirror Merge Intervals question difficulty definition with LeetCode link
nhantrivinh1 f1552eb
Update Swift naming convention in accordance with latest style guide …
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Original file line number | Diff line number | Diff line change |
---|---|---|
@@ -1,44 +1,29 @@ | ||
/** | ||
* Question Link: https://leetcode.com/problems/merge-intervals/ | ||
* Primary idea: Sort the original intervals and then append them one by one | ||
* Primary idea: Sort intervals by leading integer in ascending order and append based on their leading and trailing integers. | ||
* Time Complexity: O(nlogn), Space Complexity: O(n) | ||
* | ||
* Definition for an interval. | ||
* public class Interval { | ||
* public var start: Int | ||
* public var end: Int | ||
* public init(_ start: Int, _ end: Int) { | ||
* self.start = start | ||
* self.end = end | ||
* } | ||
* } | ||
*/ | ||
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class MergeIntervals { | ||
func merge(intervals: [Interval]) -> [Interval] { | ||
var result = [Interval]() | ||
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let intervals = intervals.sorted { | ||
if $0.start != $1.start { | ||
return $0.start < $1.start | ||
func merge(_ intervals: [[Int]]) -> [[Int]] { | ||
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. Your solution works but it is almost the same logic as the original one. Since it does not help improve the runtime complexity or simplify the code, I will not merge it. |
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guard intervals.count > 1 else { return intervals } | ||
let intervals = intervals.sorted { $0[0] < $1[0] } | ||
var ret: [[Int]] = [] | ||
var i = 1 | ||
var last = intervals[0] | ||
while i < intervals.endIndex { | ||
let curr = intervals[i] | ||
if curr[0] <= last[1] { | ||
if curr[1] > last[1] { | ||
last[1] = curr[1] | ||
} | ||
} else { | ||
return $0.end < $1.end | ||
ret.append(last) | ||
last = curr | ||
} | ||
guard i != intervals.endIndex - 1 else { ret.append(last); break } | ||
i += 1 | ||
} | ||
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for interval in intervals { | ||
guard let last = result.last else { | ||
result.append(interval) | ||
continue | ||
} | ||
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if last.end < interval.start { | ||
result.append(interval) | ||
} else { | ||
last.end = max(last.end, interval.end) | ||
} | ||
} | ||
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return result | ||
return ret | ||
} | ||
} | ||
} |
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Why do you need to use a local
nums