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var flattenArray = function(arr) { | |
var results = []; | |
arr.forEach(function(item) { | |
if(Array.isArray(item)) { | |
results = results.concat(flattenArray(item)); | |
} else { | |
results.push(item); | |
} | |
}); | |
return results; |
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var debounce = function(fn, time, immediate) { | |
var timeout; | |
return function() { | |
var that = this; | |
var args = arguments; | |
var later = function() { | |
if(!immediate){ | |
timeout = null; | |
func.call(that, args); |
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var traverseDOM = function(node, fn) { | |
fn(node); | |
node = node.firstChild; | |
while(node) { | |
traverseDOM(node, fn); | |
node = node.nextSibling; | |
} | |
} | |
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var combinations = function(str) { | |
var arr = str.split(''); | |
for( var i=0; i< arr.length; i++){ | |
genComb(arr, i+1); | |
} | |
} | |
function genComb(arr, length, str, index) { | |
str = str || ''; | |
index = index || 0; |
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/*Given a non-empty list of words, return the k most frequent elements. | |
Your answer should be sorted by frequency from highest to lowest. If two words have the same frequency, then the word with the lower alphabetical order comes first. | |
Example 1: | |
Input: ["i", "love", "leetcode", "i", "love", "coding"], k = 2 | |
Output: ["i", "love"] | |
Explanation: "i" and "love" are the two most frequent words. | |
Note that "i" comes before "love" due to a lower alphabetical order. | |
Example 2: |
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