LeetCode第二天&第三天
LeetCode第二天&第三天
郭耀华
发布于 2018-05-09 11:24:58
发布于 2018-05-09 11:24:58
leetcode 第二天
2017年12月27日
4.（118）Pascal's Triangle
JAVA
class Solution {
public List<List<Integer>> generate(int numRows) {
List<List<Integer>> triangle = new ArrayList<List<Integer>>();
//当numRows = 0 的情况
if(numRows == 0) return triangle;
//当numRows != 0 的情况
triangle.add(new ArrayList<Integer>());
triangle.get(0).add(1);
for(int i = 1; i <numRows;i++){
List<Integer> curRow = new ArrayList<Integer>();
List<Integer> preRow = triangle.get(i-1);
//first element
curRow.add(1);
for(int j = 0 ;j<preRow.size()-1;j++){
curRow.add(preRow.get(j)+preRow.get(j+1));
}
curRow.add(1);
triangle.add(curRow);
}
return triangle;
}
}Python
def generate(self, numRows):
"""
:type numRows: int
:rtype: List[List[int]]
"""
triangle = []
if numRows == 0:
return triangle
triangle.append([1])
for i in range(1,numRows):
curRow = []
preRow = triangle[i-1]
curRow.append(1)
for j in range(len(preRow)-1):
curRow.append(preRow[j]+preRow[j+1])
curRow.append(1)
triangle.append(curRow)
return triangleleetcode 第三天
2018年1月3日
5.（217）Contains Duplicate
JAVA
- // T:O(nlogn) S:O(1) public boolean containsDuplicate(int[] nums) { Arrays.sort(nums); for (int i = 0; i < nums.length - 1; ++i) { if (nums[i] == nums[i + 1]) return true; } return false; }
- T:O(n) S:O(n) public boolean containsDuplicate(int[] nums) { Set<Integer> set = new HashSet<>(nums.length); for (int x: nums) { if (set.contains(x)) return true; set.add(x); } return false; } 6.（717）1-bit and 2-bit Characters
JAVA
- class Solution { public boolean isOneBitCharacter(int[] bits) { int i = 0; while (i<bits.length - 1){ i += bits[i]+1; } return i == bits.length-1; } }
class Solution {
public boolean isOneBitCharacter(int[] bits) {
int i = bits.length - 2;
while (i >= 0 && bits[i] > 0) i--;
return (bits.length - i) % 2 == 0;
}
}Python
- class Solution(object): def isOneBitCharacter(self, bits): i = 0 while i < len(bits) - 1: i += bits[i] + 1 return i == len(bits) - 1
class Solution(object):
def isOneBitCharacter(self, bits):
parity = bits.pop()
while bits and bits.pop(): parity ^= 1
return parity == 07.（119）Pascal's Triangle II
Java
class Solution {
public List<Integer> getRow(int rowIndex) {
List<Integer> result = new ArrayList<Integer>();
if(rowIndex < 0) return result;
for(int i = 0 ;i <= rowIndex;i++){
result.add(1);
for(int j = i-1 ; j > 0 ; j--){
result.set(j,result.get(j)+result.get(j-1));
}
}
return result;
}
}8.（695）Max Area of Island
JAVA
递归
class Solution {
int[][] grid;
boolean[][] seen;
public int maxAreaOfIsland(int[][] grid) {
this.grid = grid;
int result = 0;
seen = new boolean[grid.length][grid[0].length];
for(int r = 0;r<grid.length;r++)
for(int c = 0 ;c<grid[0].length;c++)
result = Math.max(result,area(r,c));
return result;
}
public int area(int r,int c){
if(r<0||r>=grid.length||c<0||c>=grid[0].length||seen[r][c]||grid[r][c]==0) return 0;
seen[r][c] = true;
return 1+area(r-1,c)+area(r,c-1)+area(r+1,c)+area(r,c+1);
}
}9.（26）Remove Duplicates from Sorted Array
JAVA
class Solution {
public int removeDuplicates(int[] nums) {
int newLength = 1;
if(nums.length == 0) return 0;
for(int i = 0;i<nums.length;i++)
if(nums[newLength-1] != nums[i]){
nums[newLength]= nums[i];
newLength++;
}
return newLength;
}
}10.（27）Remove Element
JAVA
class Solution {
public int removeElement(int[] nums, int val) {
int newLength = 0;
if(nums.length ==0) return newLength;
for(int i = 0;i<nums.length;i++)
if(nums[i]!=val)
nums[newLength++] = nums[i];
return newLength;
}
}11.（121）Best Time to Buy and Sell Stock
JAVA
class Solution {
public int maxProfit(int[] prices) {
int profit = 0;
int min = Integer.MAX_VALUE;
for(int i = 0;i<prices.length;i++){
min = Math.min(prices[i],min);
profit = Math.max(prices[i]-min,profit);
}
return profit;
}
}12.（122）Best Time to Buy and Sell Stock II
JAVA
class Solution {
public int maxProfit(int[] prices) {
int profit = 0;
for(int i =0;i<prices.length-1;i++)
if(prices[i+1]>prices[i])
profit += prices[i+1]-prices[i];
return profit;
}
}13.（624）==Maximum Distance in Arrays==
Given m arrays, and each array is sorted in ascending order. Now you can pick up two integers from two different arrays (each array picks one) and calculate the distance. We define the distance between two integers a and b to be their absolute difference |a-b|. Your task is to find the maximum distance.
Example 1:
Input: [[1,2,3], [4,5], [1,2,3]] Output: 4 Explanation: One way to reach the maximum distance 4 is to pick 1 in the first or third array and pick 5 in the second array.
Note:
- Each given array will have at least 1 number. There will be at least two non-empty arrays.
- The total number of the integers in all the m arrays will be in the range of [2, 10000].
- The integers in the m arrays will be in the range of [-10000, 10000]. JAVA
public class Solution {
public int maxDistance(List<List<Integer>> arrays) {
int res = 0;
int min = arrays.get(0).get(0);
int max = arrays.get(0).get(arrays.get(0).size() - 1);
for (int i = 1; i < arrays.size(); i++) {
List<Integer> array = arrays.get(i);
res = Math.max(Math.abs(min - array.get(array.size() - 1)), Math.max(Math.abs(array.get(0) - max), res));
min = Math.min(min, array.get(0));
max = Math.max(max, array.get(array.size() - 1));
}
return res;
}
} 14.（35）Search Insert Position
JAVA
class Solution {
public int searchInsert(int[] nums, int target) {
for(int i =0;i<nums.length;i++){
if(nums[i] == target)
return i;
if(nums[i]>target)
return i;
}
return nums.length;
}
}本文参与 腾讯云自媒体分享计划，分享自作者个人站点/博客。
原始发表：2018-01-03 ，如有侵权请联系 cloudcommunity@tencent.com 删除
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