标签：++ ide length type code list pen str 新规
1.Two Sum
Given an array of integers, return indices of the two numbers such that they add up to a specific target. You may assume that each input would have exactly one solution, and you may not use the same element twice.
使用hash
public int[] twoSum(int[] nums, int target) { int[] res = new int[2]; if (nums == null || nums.length < 2) { return res; } HashMap<Integer, Integer> map = new HashMap<Integer, Integer>(); int n = nums.length; for (int i = 0; i < n; i++) { if (map.containsKey(target - nums[i])) { res[0] = map.get(target - nums[i]); res[1] = i; return res; } map.put(nums[i], i); } return res; }
2.Add Two Numbers
You are given two non-empty linked lists representing two non-negative integers. 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. You may assume the two numbers do not contain any leading zero, except the number 0 itself. Input: (2 -> 4 -> 3) + (5 -> 6 -> 4) Output: 7 -> 0 -> 8
dummyHead cur控制。carry携带进位
def addTwoNumbers(self, l1, l2): """ :type l1: ListNode :type l2: ListNode :rtype: ListNode """ dummyHead = ListNode(0) cur = dummyHead carry = 0 while (l1 or l2): sum = 0 if l1: sum += l1.val l1 = l1.next if l2: sum += l2.val l2 = l2.next sum += carry carry = sum / 10 cur.next = ListNode(sum % 10) cur = cur.next if carry != 0: cur.next = ListNode(carry) return dummyHead.next
3.Longest Substring Without Repeating Characters --- not bug free
Given a string, find the length of the longest substring without repeating characters. Examples: 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.
使用数组充当map
注意：left的更新规则，可能有些出现过的点上次出现的位置还在left的左边。最后也需要一个max的操作
1 public int lengthOfLongestSubstring(String s) { 2 if (s == null || s.length() == 0) { 3 return 0; 4 } 5 int[] map = new int[128]; 6 Arrays.fill(map, -1); 7 char[] chars = s.toCharArray(); 8 int n = s.length(); 9 int left = 0; 10 int right = 0; 11 int res = 0; 12 while (right < n) { 13 if (map[chars[right]] == -1) { 14 map[chars[right]] = right; 15 right++; 16 } else { 17 res = Math.max(res, right - left); 18 left = Math.max(left, map[chars[right]] + 1); // Attention! 19 map[chars[right]] = right; 20 right++; 21 } 22 } 23 res = Math.max(res, right - left); 24 return res; // Attention! 25 }
4.Median of Two Sorted Arrays
There are two sorted arrays nums1 and nums2 of size m and n respectively. Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)).
findTopK。检测四个条件：是否一个为空了，是否找第一个，是否其中一个不够 k / 2 了，谁大谁小。
public double findMedianSortedArrays(int[] nums1, int[] nums2) { if (nums1 == null || nums2 == null) { return 0.0; } int m = nums1.length; int n = nums2.length; int total = m + n; if (total % 2 == 0) { return (findTopK(nums1, 0, m - 1, nums2, 0, n - 1, total / 2) + findTopK(nums1, 0, m - 1, nums2, 0, n - 1, total / 2 + 1)) / 2.0; } else { return (double)findTopK(nums1, 0, m -1, nums2, 0, n - 1, total / 2 + 1); } } public int findTopK(int[] nums1, int start1, int end1, int[] nums2, int start2, int end2, int k) { if (start1 > end1) { return nums2[start2 + k - 1]; } if (start2 > end2) { return nums1[start1 + k - 1]; } if (k == 1) { return Math.min(nums1[start1], nums2[start2]); } if (start1 + k / 2 - 1 > end1) { return findTopK(nums1, start1, end1, nums2, start2 + k / 2, end2, k - k / 2); } if (start2 + k / 2 - 1 > end2) { return findTopK(nums1, start1 + k / 2, end1, nums2, start2, end2, k - k / 2); } if (nums1[start1 + k / 2 - 1] > nums2[start2 + k / 2 - 1]) { return findTopK(nums1, start1, end1, nums2, start2 + k / 2, end2, k - k / 2); } else { return findTopK(nums1, start1 + k / 2, end1, nums2, start2, end2, k - k / 2); } }
标签：++ ide length type code list pen str 新规
原文地址：http://www.cnblogs.com/futurehau/p/6564471.html
