2、位元組跳動-陣列與排序
阿新 • • 發佈：2018-12-10
1、三數之和
class Solution { public List<List<Integer>> threeSum(int[] nums) { List<List<Integer>> result = new ArrayList<>(); Arrays.sort(nums); for (int i = 0; i < nums.length; i++) { if (nums[i] > 0) break; if (i > 0 && nums[i] == nums[i - 1]) continue; int j = nums.length - 1; int target = 0 - nums[i]; int k = i + 1; while (k < j) { if (nums[k] + nums[j] == target) { List<Integer> item = Arrays.asList(nums[i], nums[k], nums[j]); result.add(item); while (k < j && nums[k] == nums[k + 1]) k++; while (k < j && nums[j] == nums[j - 1]) j--; k++;j--; } else if (nums[k] + nums[j] < target) { k++; } else { j--; } } } return result; } }
原文：https://blog.csdn.net/qq_35170267/article/details/81031368
2、島嶼的最大面積
class Solution { public int maxAreaOfIsland(int[][] grid) { int i,j,temp,result=0; for(i=0;i<grid.length;i++){ for(j=0;j<grid[0].length;j++){ if(grid[i][j]==1){ result=Math.max((temp=countArea(grid,i,j)),result); } } } return result; } public int countArea(int [][] grid,int i,int j){ if(i<0||j<0||i>=grid.length||j>=grid[0].length||grid[i][j]!=1) return 0; grid[i][j]=0; return 1+countArea(grid,i+1,j)+countArea(grid,i-1,j)+countArea(grid,i,j+1)+countArea(grid,i,j-1); } }
原文：https://blog.csdn.net/qq_38959715/article/details/80937405
3、搜尋旋轉排序陣列
public int search(int[] nums, int target) { if(nums==null||nums.length<1) return -1; int left = 0; int right = nums.length - 1; while (left <= right) { int mid = left + (right - left) / 2; if (nums[mid] == target) return mid; //條件1 if (nums[mid] >= nums[left]) { if (target < nums[mid] && target >= nums[left]) { right = mid - 1; }else { left = mid + 1; } } //條件2 if (nums[mid] <= nums[right]) { if (target > nums[mid] && target <= nums[right]) { left = mid + 1; }else { right = mid - 1; } } } return -1; }
參考：https://www.cnblogs.com/keeya/p/9689927.html
4、最長連續遞增序列
public class Test48 {
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner sc = new Scanner(System.in);
int n=sc.nextInt();
int[] res=new int[n];
for(int i=0;i<n;i++)
res[i]=sc.nextInt();
System.out.println(getMaxlength(res));
}
public static int getMaxlength(int[] A) {
int size=A.length;
if (size <= 0)
return 0;
int res = 1;
int current = 1;
for (int i = 1; i < size; i++) {
if (A[i] > A[i - 1]) {
current++;
} else {
if (current > res) {
res = current;
}
current = 1;
}
}
return res;
}
}參考：https://blog.csdn.net/boguesfei/article/details/82901414
5、陣列中的第K個最大元素
public int findKthLargest(int[] nums, int k) {
Arrays.sort(nums);
return nums[nums.length - k];
}參考：https://blog.csdn.net/ccccc1997/article/details/81673753
6、最長連續序列
public int longestcontinueArrays(int arr[])
{
if(arr==null||arr.length==0)
return 0;
int longest=0;
int len=1;
Arrays.sort(arr);
//對陣列進行排序
for(int i=0;i<arr.length;i++)
{
if(i+1<arr.length&&arr[i]+1==arr[i+1])
{
len++;
longest=Math.max(longest, len);
//longest 儲存當前最長連續陣列的長度。
}
else {
len=1;
//當不連續時將len=1
}
}
return longest;
}原文：https://blog.csdn.net/u013309870/article/details/70242770
7、第k個排列
public class Solution {
public String getPermutation(int n, int k) {
k--;
List<Integer> list = new ArrayList<Integer>();//注意儲存1-n
StringBuilder s = new StringBuilder();
int times = n-1;
for(int i=1;i<=n;i++){
list.add(i);
}
int factorail = 1;//階乘
for(int i=2;i<n;i++){//不要×n
factorail*=i;
}
while(times>=0){
int indexList = k/factorail;
s.append(list.get(indexList));
list.remove(indexList);
k=k%factorail;
if(times!=0){
factorail/=times;
}
times--;
}
return s.toString();
}
}
原文：https://blog.csdn.net/Lynn_Baby/article/details/80948414
8、朋友圈
public class Solution {
public void dfs(int[][] M, int[] visited, int i) {
for (int j = 0; j < M.length; j++) {
if (M[i][j] == 1 && visited[j] == 0) {
visited[j] = 1;
dfs(M, visited, j);
}
}
}
public int findCircleNum(int[][] M) {
int[] visited = new int[M.length];
int count = 0;
for (int i = 0; i < M.length; i++) {
if (visited[i] == 0) {
dfs(M, visited, i);
count++;
}
}
return count;
}
}原文：https://blog.csdn.net/mine_song/article/details/70195463
9、合併區間
public class Solution {
public List<Interval> merge(List<Interval> intervals) {
List<Interval> result = new LinkedList<>();
if (intervals == null || intervals.size() < 1) {
return result;
}
// 先對區間進行排序，使用一個匿名內部類
Collections.sort(intervals, new Comparator<Interval>() {
@Override
public int compare(Interval o1, Interval o2) {
return o1.start - o2.start;
}
});
// 排序後，後一個元素（記為next）的start一定是不小於前一個（記為prev）start的，
// 對於新加入的區間，假設next.start大於prev.end就說明這兩個區間是分開的，要添
// 加一個新的區間。否則說明next.start在[prev.start, prev.end]內。則僅僅要看
// next.end是否是大於prev.end，假設大於就要合併區間（擴大）
Interval prev = null;
for (Interval item : intervals) {
if (prev == null || prev.end < item.start) {
result.add(item);
prev = item;
} else if (prev.end < item.end) {
prev.end = item.end;
}
}
return result;
}
}參考：https://www.cnblogs.com/gccbuaa/p/7088508.html
10、接雨水
public class Solution {
/**
* @param heights: an array of integers
* @return: a integer
*/
public int trapRainWater(int[] heights) {
// write your code here
int left = 0, right = heights.length - 1;
int res = 0;
if(left >= right)
return res;
int leftheight = heights[left];
int rightheight = heights[right];
while(left < right) {
if(leftheight < rightheight) {
left ++;
if(leftheight > heights[left]) {
res += (leftheight - heights[left]);
} else {
leftheight = heights[left];
}
} else {
right --;
if(rightheight > heights[right]) {
res += (rightheight - heights[right]);
} else {
rightheight = heights[right];
}
}
}
return res;
}
}原文：https://blog.csdn.net/qq_14927217/article/details/72861208