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+ /**
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+ * @description
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+ * 인덱스로 접근하지 못하는 구조를 인덱스로 접근하게 하여 two pointer로 풀이
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+ *
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+ * n = total node count
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+ * time complexity: O(n)
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+ * space complexity: O(n)
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+ */
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+ var reorderList = function ( head ) {
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+ // convert from queue to list
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+ let travelNode = head ;
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+ const list = [ ] ;
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+ while ( travelNode ) {
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+ list . push ( travelNode ) ;
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+ travelNode = travelNode . next ;
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+ }
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+ // two pointer
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+ let [ left , right ] = [ 0 , list . length - 1 ] ;
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+ const node = new ListNode ( ) ;
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+ let tail = node ;
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+
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+ while ( left <= right ) {
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+ // 1. left append
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+ const leftNode = list [ left ] ;
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+ leftNode . next = null ;
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+ tail . next = leftNode ;
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+ tail = leftNode ;
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+ // 2. conditional right append
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+ const rightNode = list [ right ] ;
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+ rightNode . next = null ;
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+ if ( left !== right ) {
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+ tail . next = rightNode ;
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+ tail = rightNode ;
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+ }
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+
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+ left ++ ;
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+ right -- ;
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+ }
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+
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+ head = node . next ;
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+ } ;
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