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Commitd756a2b

Browse files
committed
160, 19, 190, 205, 219, 223
1 parentfcb4c87 commitd756a2b

7 files changed

+182
-0
lines changed

‎Easy/111-minimumDepthBinaryTree.js‎

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* }
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*/
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/**
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* very fast 111ms
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*@param {TreeNode} root
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*@return {number}
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*/
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/**
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* Definition for singly-linked list.
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* function ListNode(val) {
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* this.val = val;
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* this.next = null;
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* }
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*/
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/**
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* because there is an intersection btw two linkedlists, first align the head of longer list with
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* the shorter one's head. Then just keep moving the heads of both linkedlists
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*@param {ListNode} headA
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*@param {ListNode} headB
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*@return {ListNode}
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*/
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vargetIntersectionNode=function(headA,headB){
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varlenA=getLength(headA);
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varlenB=getLength(headB);
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while(lenA<lenB){
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headB=headB.next;
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lenB--;
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}
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while(lenB<lenA){
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headA=headA.next;
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lenA--;
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}
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while(headA!==headB){
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headA=headA.next;
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headB=headB.next;
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}
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returnheadA;
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};
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vargetLength=function(listHead){
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varlength=0;
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while(listHead){
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length++;
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listHead=listHead.next;
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}
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returnlength;
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};
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/**
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* Definition for singly-linked list.
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* function ListNode(val) {
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* this.val = val;
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* this.next = null;
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* }
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*/
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/**
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* key: Two pointers, the distance of two pointer is n, when the second
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* pointer reaches to the end, the first pointer node is the node to be
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* deleted. Be careful when the second pointer reaches the end, first pointer
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* has not moved, then just return the second node as the new head.
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*
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*@param {ListNode} head
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*@param {number} n
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*@return {ListNode}
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*/
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varremoveNthFromEnd=function(head,n){
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varfHead=head,sHead=head;
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vardistance=0;
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while(distance<n){
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sHead=sHead.next;
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distance++;
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}
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if(!sHead)returnfHead.next;
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while(sHead.next){
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fHead=fHead.next;
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sHead=sHead.next;
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}
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fHead.next=fHead.next.next;
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returnhead;
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};

‎Easy/190-reverseBits.js‎

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/**
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* the key is to use bit AND &, after AND 1, only the last bit is reseved.
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*@param {number} n - a positive integer
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*@return {number} - a positive integer
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*/
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varreverseBits=function(n){
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varresult=0;
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for(vari=0;i<32;i++){
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varlastBit=n&1;
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result+=lastBit*Math.pow(2,31-i);
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n>>=1;
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}
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returnresult;
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};

‎Easy/205-isomorphicStrings.js‎

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/**
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* A simply solution but not efficient consider every character of ‘str1′ and
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* check if all occurrences of it map to same character occurrences in ‘str2′.
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* Time complexity of this solution is O(n*n).
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*
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* This is a better solution (NOT ACCEPTED, Time limit exceeded).
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* 1. check if t[i] is in the map. If yes, find out the key associated with t[i]
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* if key exists, but s[i] is not equal to key, return false;
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* if key is null, but s[i] exists in the map and map[s[i]] is not equal to t[i]
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* return false.
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*
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*@param {string} s
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*@param {string} t
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*@return {boolean}
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*/
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varisIsomorphic=function(s,t){
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varmap={};
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for(vari=0;i<s.length;i++){
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varkey=map.getKeyByValue(t[i]);
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if(key&&s[i]!==key){
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returnfalse;
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}elseif(s[i]inmap){
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if(map[s[i]]!==t[i]){
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returnfalse;
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}
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}else{
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map[s[i]]=t[i];
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}
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}
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returntrue;
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};
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Object.prototype.getKeyByValue=function(value){
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for(varpropinthis){
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if(this.hasOwnProperty(prop)){
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if(this[prop]===value)
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returnprop;
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}
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}
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};

‎Easy/219-containsDuplicateII.js‎

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/**
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* key: use JavaScript Object to keep track the num and its index
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* REMEBER: use if (key in Obj) to check if the Obj has the key 'key'
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*@param {number[]} nums
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*@param {number} k
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*@return {boolean}
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*/
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varcontainsNearbyDuplicate=function(nums,k){
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varnumIndex={};
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for(vari=0;i<nums.length;i++){
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if(nums[i]innumIndex){
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if(i-numIndex[nums[i]]<=k){
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returntrue;
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}
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}
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numIndex[nums[i]]=i;
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}
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returnfalse;
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};

‎Easy/223-rectangleArea.js‎

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/**
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* Key: find the left most x, right most x, top most y, top most y,
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* that is the overlap rectangle.
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*@param {number} A
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*@param {number} B
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*@param {number} C
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*@param {number} D
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*@param {number} E
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*@param {number} F
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*@param {number} G
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*@param {number} H
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*@return {number}
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*/
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varcomputeArea=function(A,B,C,D,E,F,G,H){
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varareaA=(C-A)*(D-B);
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varareaB=(G-E)*(H-F);
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varleft=Math.max(A,E);
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varright=Math.min(C,G);
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vartop=Math.min(D,H);
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varbottom=Math.max(B,F);
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varoverlapArea=0;
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if(right>left&&top>bottom){
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overlapArea=(right-left)*(top-bottom);
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}
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returnareaA+areaB-overlapArea;
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};

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