Don't say so much, just coding...
Write a function called sumIntervals/sum_intervals() that accepts an array of intervals, and returns the sum of all the interval lengths. Overlapping intervals should only be counted once.
Intervals
Intervals are represented by a pair of integers in the form of an array. The first value of the interval will always be less than the second value. Interval example: [1, 5] is an interval from 1 to 5. The length of this interval is 4.
Overlapping Intervals
List containing overlapping intervals:
[
[1,4],
[7, 10],
[3, 5]
]
The sum of the lengths of these intervals is 7. Since [1, 4] and [3, 5] overlap, we can treat the interval as [1, 5], which has a length of 4.
Examples:
sumIntervals( [
[1,2],
[6, 10],
[11, 15]
] ); // => 9
sumIntervals( [
[1,4],
[7, 10],
[3, 5]
] ); // => 7
sumIntervals( [
[1,5],
[10, 20],
[1, 6],
[16, 19],
[5, 11]
] ); // => 19
def sum_of_intervals(intervals)
#return
end
Test.assert_equals(sum_of_intervals([[1, 5]]), 4)
Test.assert_equals(sum_of_intervals([[1, 5], [6, 10]]), 8)
Test.assert_equals(sum_of_intervals([[1, 5], [1, 5]]), 4)
Test.assert_equals(sum_of_intervals([[1, 4], [7, 10], [3, 5]]), 7)
function sumIntervals(intervals){
//TODO
}
describe('sumIntervals', function(){
it('should return the correct sum for non overlapping intervals', function(){
var test1 = [[1,5]];
var test2 = [[1,5],[6,10]];
Test.assertEquals(sumIntervals(test1), 4);
Test.assertEquals(sumIntervals(test2), 8);
});
it('should return the correct sum for overlapping intervals', function(){
var test1 = [[1,5],[1,5]];
var test2 = [[1,4],[7, 10],[3, 5]];
Test.assertEquals(sumIntervals(test1), 4);
Test.assertEquals(sumIntervals(test2), 7);
});
});
想法(1): 第一个直觉都先 map
来处理,但原来发现有 flat_map
可以更清楚来做
想法(2): 注意要为 uniq
才是对的
# Solution 1
def sum_of_intervals(intervals)
intervals.map{ |a| (a[0]...a[1]).to_a }.flatten.uniq.size
end
# Solution 2
def sum_of_intervals(intervals)
intervals.flat_map { |x, y| [*x...y] }.uniq.size
end
// Solution 1
function sumIntervals(intervals){
var numbers = {};
intervals.forEach(function(x) {
for (var i = x[0]; i < x[1]; i++) {
numbers[i] = i;
}
});
return Object.keys(numbers).length;
}
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