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Commit0736064

Browse files
committed
add some modification and new problem
1 parent75ce46f commit0736064

7 files changed

+209
-24
lines changed

‎2 Add Two Numbers.js

Lines changed: 11 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -1,3 +1,9 @@
1+
// You are given two linked lists representing two non-negative numbers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.
2+
3+
// Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
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// Output: 7 -> 0 -> 8
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6+
17
/**
28
* Definition for singly-linked list.
39
* function ListNode(val) {
@@ -10,6 +16,11 @@
1016
*@param {ListNode} l2
1117
*@return {ListNode}
1218
*/
19+
20+
21+
22+
// value reverse helps to asslign the first digit or both linklist
23+
1324
varaddTwoNumbers=function(l1,l2){
1425
if(l1===null||l2===null){
1526
returnl1||l2;

‎3 Longest Substring Without Repeating Characters.js

Lines changed: 7 additions & 4 deletions
Original file line numberDiff line numberDiff line change
@@ -2,6 +2,8 @@
22
*@param {string} s
33
*@return {number}
44
*/
5+
6+
57
varlengthOfLongestSubstring=function(s){
68
if(s===null||s.length===0){
79
return0;
@@ -11,16 +13,17 @@ var lengthOfLongestSubstring = function(s) {
1113
varlen=0;
1214
varmaxLen=len;
1315
varstart=0;
14-
16+
17+
// scan from left to right.
1518
for(vari=start;i<s.length;i++){
1619
c=s[i];
1720

1821
if(map[c]!==undefined&&map[c]>=start){
19-
start=map[c]+1;
20-
len=i-start;
22+
start=map[c]+1;// start new search with repeated word's last location + 1
23+
len=i-start;// real length -> from for example 3 to 5 is 3, so it's 5-3+1 and + 1 happens later
2124
}
2225

23-
len++;
26+
len++;// real length -> from for example 3 to 5 is 3, so it's 5-3+1 and + 1 happens later
2427

2528
if(len>maxLen){
2629
maxLen=len;

‎5 Longest Palindromic Substring.js

Lines changed: 58 additions & 5 deletions
Original file line numberDiff line numberDiff line change
@@ -1,23 +1,34 @@
1+
// Given a string S, find the longest palindromic substring in S. You may assume that the maximum length of S is 1000, and there exists one unique longest palindromic substring.
2+
13
/**
24
*@param {string} s
35
*@return {string}
46
*/
57

8+
9+
610
varlongestPalindrome=function(s){
711
if(s===null||s.length===0){
812
return"";
913
}
1014

1115
varresult="";
16+
17+
// The reason to multiply by 2 is because
18+
// Palindrome can be in two forms
19+
// 1. abba There will be case which center has two idenitical charachter,
20+
// And there will be maximum 2*n - 1 such case
21+
// 2. aba There will be case which center has only one character
22+
varlen=2*s.length-1;
23+
varleft,right;
1224

13-
for(vari=(2*s.length)-1;i--;){
14-
varleft=parseInt(i/2);
15-
varright=parseInt(i/2);
25+
for(vari=0;i<len;i++){
26+
left=right=parseInt(i/2);
1627
if(i%2===1){
1728
right++;
1829
}
1930

20-
varstr=lengthOfPalindrome(s,left,right);
31+
varstr=expandFromCenterAndCheckForPalindrome(s,left,right);
2132

2233
if(str.length>result.length){
2334
result=str;
@@ -27,7 +38,49 @@ var longestPalindrome = function(s) {
2738
};
2839

2940

30-
varlengthOfPalindrome=function(s,left,right){
41+
// other implementation
42+
43+
varlongestPalindrome=function(s){
44+
if(s===null||s.length===0){
45+
return"";
46+
}
47+
48+
varresult="";
49+
50+
// The reason to multiply by 2 is because
51+
// Palindrome can be in two forms
52+
// 1. abba There will be case which center has two idenitical charachter,
53+
// And there will be maximum 2*n - 1 such case
54+
// 2. aba There will be case which center has only one character
55+
varlen=s.length;
56+
varleft,right;
57+
58+
for(vari=0;i<len;i++){
59+
left=right=i;
60+
61+
varstr=expandFromCenterAndCheckForPalindrome(s,left,right);
62+
if(str.length>result.length){
63+
result=str;
64+
}
65+
varstr=expandFromCenterAndCheckForPalindrome(s,left,right+1);
66+
if(str.length>result.length){
67+
result=str;
68+
}
69+
}
70+
returnresult;
71+
};
72+
73+
74+
75+
varexpandFromCenterAndCheckForPalindrome=function(s,left,right){
76+
// in the case where left !== right
77+
// that's the case "abba"
78+
// which it checks for if b === b then a === a
79+
80+
// in the case where left === right
81+
// that's the case "aba"
82+
// which it check if b === b as left === right
83+
// then a === a
3184
while(left>=0&&right<s.length&&s[left]===s[right]){
3285
left--;
3386
right++;

‎9 Palindrome Number.js

Lines changed: 24 additions & 15 deletions
Original file line numberDiff line numberDiff line change
@@ -2,35 +2,44 @@
22
// Language: Javascript
33
// Problem: https://leetcode.com/problems/palindrome-number/
44
// Author: Chihung Yu
5-
/**
6-
* Leetcode Question
5+
6+
// Determine whether an integer is a palindrome. Do this without extra space.
7+
8+
/**
79
*@param {number} x
810
*@return {boolean}
911
*/
10-
1112
varisPalindrome=function(x){
12-
if(x<0){
13+
if(x<0){
1314
returnfalse;
1415
}
1516

1617
vardiv=1;
1718

18-
while(parseInt(x/div)>=10){
19-
div*=10;
19+
// it means that the div is still a valid divider
20+
// e.g 600 the divider should be 100 at maximum
21+
// e.g. 90 the divider should be 10 at maximum
22+
// e.g. 1 the divider should be a 1 at maximum
23+
while(parseInt(x/div)>=10){
24+
div*=10;
2025
}
2126

22-
while(x!==0){
23-
varl=parseInt(x/div);
24-
varr=x%10;
25-
26-
if(l!==r){
27+
varleft,right;
28+
29+
// when div === 1 it means the digit only has one value to examine
30+
// e.g. 121 -> only 2 is left for examine which can be ignore
31+
// e.g. 1 -> 1
32+
// e.g. 43234 -> 2
33+
while(div>1){
34+
left=parseInt(x/div);
35+
right=x%10;
36+
if(left!==right){
2737
returnfalse;
2838
}
2939

30-
x=x%div;
31-
x=parseInt(x/10);
32-
33-
div=parseInt(div/100);
40+
x=x%div;// remove the left most digit
41+
x=parseInt(x/10);// remove the right most digit
42+
div/=100;
3443
}
3544

3645
returntrue;

‎Jump Game II.js

Lines changed: 25 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,25 @@
1+
/**
2+
*@param {number[]} nums
3+
*@return {number}
4+
*/
5+
varjump=function(nums){
6+
varcurMax=0;
7+
varnJumps=0;
8+
vari=0;
9+
varn=nums.length;
10+
11+
while(curMax<n-1){
12+
varlastMax=curMax;
13+
// go through covered area
14+
for(;i<=lastMax;i++){
15+
curMax=Math.max(curMax,i+nums[i]);
16+
}
17+
nJumps++;
18+
// if cannot make progress in the covered area, give up
19+
if(lastMax===curMax){
20+
return-1;
21+
}
22+
}
23+
24+
returnnJumps;
25+
};

‎Reverse Integer.js

Lines changed: 25 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,25 @@
1+
/**
2+
*@param {number} x
3+
*@return {number}
4+
*/
5+
varreverse=function(x){
6+
varisNeg=x<0;
7+
varresult=0;
8+
9+
x=Math.abs(x);
10+
11+
while(x){
12+
varlastDigit=x%10;
13+
result*=10;
14+
result+=lastDigit;
15+
x=parseInt(x/10);
16+
}
17+
18+
result=isNeg ?-result :result;
19+
20+
if(result>Math.pow(2,31)-1||result<-Math.pow(2,31)){
21+
return0;
22+
}
23+
24+
returnresult;
25+
};

‎find kth element in two arrays.js

Lines changed: 59 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,59 @@
1+
varfindKth=(a1,a2,kth)=>{
2+
varalen1=a1.length;
3+
varalen2=a2.length;
4+
5+
if(alen1>alen2){
6+
returnfindKth(a2,a1,kth);
7+
}
8+
9+
if(alen1===0){
10+
returna2[kth-1];
11+
}
12+
13+
if(kth===1){
14+
returnMath.min(a1[0],a2[0]);
15+
}
16+
17+
// need to make sure that kth is not 1 as kth/2 = 0
18+
varp1=Math.min(parseInt(kth/2),alen1);
19+
varp2=kth-p1;
20+
21+
if(a1[p1-1]<a2[p2-1]){
22+
returnfindKth(a1.slice(p1),a2,kth-p1);
23+
}elseif(a1[p1-1]>a2[p2-1]){
24+
returnfindKth(a1,a2.slice(p2),kth-p2);
25+
}else{
26+
returna1[p1-1];
27+
}
28+
}
29+
30+
31+
// function findKth(a, aStart, aEnd, b, bStart, bEnd, kth) {
32+
// // use array position rather than use slice to get array
33+
// // always assume that m is equal or smaller than n
34+
// var aLen = aEnd - aStart + 1; // same as a.length as index 3 - 3 = 0 but still it has one element compared to a.length === 0 has notthing
35+
// var bLen = bEnd - bStart + 1;
36+
// var mid;
37+
// if(aLen > bLen) {
38+
// return findKth(b, bStart, bEnd, a, aStart, aEnd, kth);
39+
// }
40+
41+
// if(aLen <= 0) {
42+
// return b[kth - 1];
43+
// }
44+
45+
// if(kth === 1) {
46+
// return Math.min(a[aStart], b[bStart]);
47+
// }
48+
49+
// var k1 = Math.min(parseInt(kth/2), aLen);
50+
// var k2 = kth - k1;
51+
52+
// if(a[aStart + k1 - 1] < b[bStart + k2 - 1]) {
53+
// return findKth(a, aStart + k1, aEnd, b, bStart, bEnd, kth -k1);
54+
// } else if(a[aStart + k1 - 1] > b[bStart + k2 - 1]) {
55+
// return findKth(a, aStart, aEnd, b, bStart + k2, bEnd, kth -k2);
56+
// } else {
57+
// return a[aStart + k1 -1];
58+
// }
59+
// }

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