K-diff Pairs in an Array leetcode 532
2017-03-07 08:22
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题目大意:给定一个数组,从中取出差值绝对值为k的pair,pair不能重复。
题目分析:如果调用combination之类的函数,会造成实际上的算法复杂度为O(n^2),最后导致了TLE。因此改为直接统计数字在数组中出现的次数,然后根据 k 的值来进行不同的判断统计。
AC code(Ruby):
def find_pairs(nums, k)
if k < 0 || nums.length < 2
return 0
end
count = 0
h = Hash.new
nums.each {|n| h
= (h.include? n) ? h
+ 1 : 1 }
if k == 0
h.each_value {|v| count += 1 if v > 1 }
elsif k > 0
h.each_key {|key| count += 1 if h.include? key + k }
end
count
end
题目分析:如果调用combination之类的函数,会造成实际上的算法复杂度为O(n^2),最后导致了TLE。因此改为直接统计数字在数组中出现的次数,然后根据 k 的值来进行不同的判断统计。
AC code(Ruby):
def find_pairs(nums, k)
if k < 0 || nums.length < 2
return 0
end
count = 0
h = Hash.new
nums.each {|n| h
= (h.include? n) ? h
+ 1 : 1 }
if k == 0
h.each_value {|v| count += 1 if v > 1 }
elsif k > 0
h.each_key {|key| count += 1 if h.include? key + k }
end
count
end
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