add more solution

Author: chihungyu1116

103 Binary Tree Zigzag Level Order Traversal.js

@@ -1,3 +1,22 @@
+// Given a binary tree, return the zigzag level order traversal of its nodes' values. (ie, from left to right, then right to left for the next level and alternate between).
+
+// For example:
+// Given binary tree [3,9,20,null,null,15,7],
+//     3
+//    / \
+//   9  20
+//     /  \
+//    15   7
+// return its zigzag level order traversal as:
+// [
+//   [3],
+//   [20,9],
+//   [15,7]
+// ]
+// Hide Company Tags LinkedIn Bloomberg Microsoft
+// Hide Tags Tree Breadth-first Search Stack
+// Hide Similar Problems (E) Binary Tree Level Order Traversal
+
 /**
  * Definition for a binary tree node.
  * function TreeNode(val) {
@@ -10,47 +29,50 @@
  * @return {number[][]}
  */
 var zigzagLevelOrder = function(root) {
-    var result = []
+    // bfs
 
     if(!root) {
-        return result;
+        return [];
     }
 
-    var fromLeft = false;
-    var curLvl = [];
-    curLvl.push(root);
-
-    var nextLvl = [];
-    var temp = [];
-
-    while(curLvl.length !== 0) {
-        var p = curLvl.pop();
-        temp.push(p.val);
+    var curLevel = [];
+    curLevel.push(root);
+    
+    var fromLeft = true;
+    var result = [];
+    var tmpResult = [];
+    var nextLevel = [];
+    
+    while(curLevel.length > 0) {
+        var len = curLevel.length;
 
-        if(fromLeft) {
-            if(p.left) {
-                nextLvl.push(p.left);
-            }
-            if(p.right) {
-                nextLvl.push(p.right);
-            }
-        } else {
-            if(p.right) {
-                nextLvl.push(p.right);
-            }
-            if(p.left) {
-                nextLvl.push(p.left);
+        for(var i = 0; i < len; i++) {
+            var node = curLevel.pop();
+            tmpResult.push(node.val);
+            
+            if(fromLeft) {
+                if(node.left) {
+                    nextLevel.push(node.left);
+                }
+                if(node.right) {
+                    nextLevel.push(node.right);
+                }
+            } else {
+                if(node.right) {
+                    nextLevel.push(node.right);
+                }
+                if(node.left) {
+                    nextLevel.push(node.left);
+                }
             }
         }
 
-        if(curLvl.length === 0) {
-            fromLeft = !fromLeft;
-            result.push(temp);
-            temp = [];
-            curLvl = nextLvl;
-            nextLvl = [];
-        }
+        fromLeft = !fromLeft;
+        curLevel = nextLevel;
+        nextLevel = [];
+        result.push(tmpResult);
+        tmpResult = [];
     }
 
-    return result
+    return result;
 };

117 Populating Next Right Pointer.js

@@ -0,0 +1,69 @@
+// Follow up for problem "Populating Next Right Pointers in Each Node".
+
+// What if the given tree could be any binary tree? Would your previous solution still work?
+
+// Note:
+
+// You may only use constant extra space.
+// For example,
+// Given the following binary tree,
+//          1
+//        /  \
+//       2    3
+//      / \    \
+//     4   5    7
+// After calling your function, the tree should look like:
+//          1 -> NULL
+//        /  \
+//       2 -> 3 -> NULL
+//      / \    \
+//     4-> 5 -> 7 -> NULL
+// Hide Company Tags Microsoft Bloomberg Facebook
+// Hide Tags Tree Depth-first Search
+// Hide Similar Problems (M) Populating Next Right Pointers in Each Node
+
+
+
+/**
+ * Definition for binary tree with next pointer.
+ * function TreeLinkNode(val) {
+ *     this.val = val;
+ *     this.left = this.right = this.next = null;
+ * }
+ */
+
+/**
+ * @param {TreeLinkNode} root
+ * @return {void} Do not return anything, modify tree in-place instead.
+ */
+var connect = function(root) {
+    if(!root) {
+        return;
+    }
+    
+    // leftEnd is used to track the current left most node
+    var leftEnd = root;
+    
+    while(leftEnd !== null) {
+        var cur = leftEnd;
+        // dummy is used to point to the next level's leftEnd
+        var dummy = new TreeLinkNode(0);
+        var pre = dummy;
+        // for each level we use leftEnd and leftEnd next to achieve level traversal
+        while(cur !== null) {
+            if(cur.left !== null) {
+                pre.next = cur.left;
+                pre = cur.left;
+            }
+            
+            if(cur.right !== null) {
+                pre.next = cur.right;
+                pre = cur.right;
+            }
+            
+            cur = cur.next;
+        }
+        
+        leftEnd = dummy.next;
+    }
+};

117 Populating Next Right Pointers in Each Node II.js

@@ -0,0 +1,40 @@
+// Given a string, determine if it is a palindrome, considering only alphanumeric characters and ignoring cases.
+
+// For example,
+// "A man, a plan, a canal: Panama" is a palindrome.
+// "race a car" is not a palindrome.
+
+// Note:
+// Have you consider that the string might be empty? This is a good question to ask during an interview.
+
+// For the purpose of this problem, we define empty string as valid palindrome.
+
+// Hide Company Tags Microsoft Uber Facebook Zenefits
+// Hide Tags Two Pointers String
+// Hide Similar Problems (E) Palindrome Linked List
+
+
+/**
+ * @param {string} s
+ * @return {boolean}
+ */
+var isPalindrome = function(s) {
+    s = s.toLowerCase();
+    var beg = 0;
+    var end = s.length - 1;
+    
+    while(beg < end) {
+        if(!s[beg].match(/[a-z0-9]/)) {
+            beg++;
+        } else if(!s[end].match(/[a-z0-9]/)) {
+            end--;
+        } else if(s[beg] !== s[end]) {
+            return false;
+        } else {
+            end--;
+            beg++;
+        }
+    }
+    
+    return true;
+};

125 Valid Palindrome.js

@@ -0,0 +1,69 @@
+解这个平面几何题有3个要点：
+
+1. 如何判断共线?
+两点成一直线，所以两点没有共线不共线之说。对于点p1(x1, y1)，p2(x2, y2)，p3(x3, y3)来说，共线的条件是p1-p2连线的斜率与p1-p3连线的斜率相同，即
+(y2-y1)/(x2-x1) = (y3-y1)/(x3-x1)
+所以对共线的n点，其中任意两点连线的斜率相同。
+
+2. 如何判断最多的共线点?
+对于每个点p出发，计算该点到所有其他点qi的斜率，对每个斜率统计有多少个点符合。其中最多的个数加1（出发点本身）即为最多的共线点。
+
+3. 特殊情况
+当x1 = x2，y1!=y2时，为垂直连线。计算斜率时分母为0会出错。
+当x1 = x2，y1 = y2时，两点重合。则(x2, y2)和所有(x1, y1)的连线共线。
+
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+class Solution {
+public:
+    int maxPoints(vector<Point> &points) {
+        int maxPts = 0;
+        for(int i=0; i<points.size(); i++) {
+            int nMax = 0, nSame = 0, nInf = 0;
+            unordered_map<float,int> comSlopes;
+            
+            for(int j=i+1; j<points.size(); j++) {
+                if(points[j].x==points[i].x) {
+                    if(points[j].y==points[i].y)
+                        nSame++;
+                    else
+                        nInf++;
+                    continue;
+                }
+                float slope = (float)(points[j].y-points[i].y)/(float)(points[j].x-points[i].x);
+                comSlopes[slope]++;
+                nMax = max(nMax, comSlopes[slope]);
+            }
+            
+            nMax = max(nMax, nInf)+nSame+1;
+            maxPts = max(maxPts,nMax);
+        }
+        return maxPts;
+    }
+};

149 Max Points on a Line.js

@@ -0,0 +1,45 @@
+// Given two strings S and T, determine if they are both one edit distance apart.
+
+// Hide Company Tags Snapchat Uber Facebook Twitter
+// Hide Tags String
+// Hide Similar Problems (H) Edit Distance
+
+
+/**
+ * @param {string} s
+ * @param {string} t
+ * @return {boolean}
+ */
+
+// tricky question!
+
+var isOneEditDistance = function(s, t) {
+  if(s.length > t.length) {
+    var tmp = s;
+    s = t;
+    t = tmp;
+  }
+
+  if((t.length - s.length) > 1) {
+    return false;
+  }
+
+  var found = false;
+  
+  for(var i = 0, j = 0; i < s.length; i++, j++) {
+    if(s[i] !== t[j]) {
+
+      if(found) {
+        return false;
+      }
+
+      found = true;
+
+      if(s.length < t.length) {
+        i--;
+      }
+    }
+  }
+
+  return found || s.length < t.length;
+};

161 One Edit Distance.js

@@ -2,6 +2,20 @@
 // Language: Javascript
 // Problem: https://leetcode.com/problems/number-of-1-bits/
 // Author: Chihung Yu
+
+// Write a function that takes an unsigned integer and returns the number of ’1' bits it has (also known as the Hamming weight).
+
+// For example, the 32-bit integer ’11' has binary representation 00000000000000000000000000001011, so the function should return 3.
+
+// Credits:
+// Special thanks to @ts for adding this problem and creating all test cases.
+
+// Hide Company Tags Microsoft Apple
+// Hide Tags Bit Manipulation
+// Hide Similar Problems (E) Reverse Bits (E) Power of Two (M) Counting Bits
+
+
+
 /**
  * @param {number} n - a positive integer
  * @return {number}