給自己的目標：[LeetCode](https://leetcode.com/ "Online Judge Platform") 上每日一題
在做題的過程中記錄下解題的思路或者重要的代碼碎片以便后來翻閱。
項目源碼：github上的Leetcode
1. Two Sum
題目：給出一個數組和一個目標值，求數組內兩個值相加與目標值相等的下標。假設唯一解。
Given nums = [2, 7, 11, 15], target = 9,
Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].
使用 HashMap 做存儲，value為num值，key為目標值減去value后需要的值。
public class Solution {
public int[] twoSum(int[] nums, int target) {
int[] answer = new int[2];
HashMap<Integer,Integer> hashMap = new HashMap<>();
for (int i=0;i<nums.length;i++){
if(hashMap.containsKey(nums[i])){
answer[0] = hashMap.get(nums[i]);
answer[1] = i;
break;
}
else{
hashMap.put(target - nums[i],i);
}
}
return answer;
}
}
2. Add Two Numbers
題目：輸入兩個鏈表，鏈表中的值一一相加，輸出鏈表。
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
由于會出現鏈表不同長和進位的情況，所以開頭要做好為空的處理。
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
public class Solution {
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
if (l1 == null || l2 == null) {
if (l1 == null && l2 == null)
return null;
if (l1 == null)
l1 = new ListNode(0);
if (l2 == null)
l2 = new ListNode(0);
}
ListNode result = new ListNode(0);
result.val = (l1.val + l2.val) % 10;
int off = (l1.val + l2.val) / 10;
if (l1.next != null) {
l1.next.val += off;
} else if ((l2.next != null)) {
l2.next.val += off;
} else if (off > 0) {
l1.next = new ListNode(off);
}
result.next = addTwoNumbers(l1.next, l2.next);
return result;
}
}
3. Longest Substring Without Repeating Characters
題目：最長不重復子串
Given "abcabcbb", the answer is "abc", which the length is 3.
Given "bbbbb", the answer is "b", with the length of 1.
Given "pwwkew", the answer is "wke", with the length of 3.
Note that the answer must be a substring, "pwke" is a subsequence and not a substring.
貪心算法，需要一個Map記錄字符最后出現的位置。同時還有一個 index 記錄不重復字符串的開頭以便計算長度和 max 最大值。
public class Solution {
public int lengthOfLongestSubstring(String s) {
int curIndex = 0;
int max = 0;
Map<Character, Integer> charMap = new HashMap<>();
for (int i = 0; i < s.length(); i++) {
if (charMap.containsKey(s.charAt(i))) {
curIndex = Math.max(charMap.get(s.charAt(i)) + 1, curIndex);
}
charMap.put(s.charAt(i), i);
max = Math.max(max, i - curIndex + 1);
}
return max;
}
}
4. Median of Two Sorted Arrays
題目：給出兩個有序數組，合并后求中心值。
nums1 = [1, 3]
nums2 = [2]
The median is 2.0
nums1 = [1, 2]
nums2 = [3, 4]
The median is (2 + 3)/2 = 2.5
給出兩個指針分別指向兩個數組的開頭，當nums1上的 數小于nums2時，nums上的指針向后移一位，一直到兩個長度和的中間值。
public class Solution {
public double findMedianSortedArrays(int[] nums1, int[] nums2) {
int m = nums1.length;
int n = nums2.length;
double sum = 0;
boolean flag = (m + n) % 2 == 1;
int median = (m + n) / 2;
int i = 0, j = 0;
int x1 = 0;
for (int k = 0; k <= n + m; k++) {
if (i < m && j < n && nums1[i] <= nums2[j]) {
x1 = nums1[i];
i++;
} else if (i < m && j < n && nums1[i] > nums2[j]) {
x1 = nums2[j];
j++;
} else if (i >= m) {
x1 = nums2[j];
j++;
} else if (j >= n) {
x1 = nums1[i];
i++;
}
if (flag) {
if (k == median) {
sum = x1 * 1.0;
break;
}
} else {
if (k == median - 1) {
sum += x1;
} else if (k == median) {
sum += x1;
sum = sum / 2.0;
break;
}
}
}
return sum;
}
}
5. Longest Palindromic Substring
題目：最長回文子串。string長度最大為1000
Input: "babad"
Output: "bab"
Note: "aba" is also a valid answer.
Input: "cbbd"
Output: "bb"
回文有兩種：（1）中間存在單個字符，如 bab；（2）左右對稱，如 bb。
所以求回文時要把兩種形式都要考慮進去。
求回文的方法：取中間一個值或中間兩個相等的值分別向左向右循環比較。遞歸求解。
public class Solution {
private int index,max;
public String longestPalindrome(String s) {
int len = s.length();
if (len < 2) return s;
for (int i = 0; i < s.length() - 1; i++) {
extendPalindrome(s, i, i);
extendPalindrome(s, i, i + 1);
}
return s.substring(index, index + max);
}
private void extendPalindrome(String s,int begin,int end){
while (begin >= 0 && end < s.length() && s.charAt(begin) == s.charAt(end)) {
begin--;
end++;
}
if (end - begin - 1 > max) {
max = end - begin - 1;
index = begin + 1;
}
}
}
