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118. Yang Hui Triangle – LeetCode
Table of contents
3. Remove duplicates in an ordered array
1. Yang Hui Triangle
Title:Given a non-negative integer numRows, generate the first part of "Yang Hui's triangle" a> numRows rows (1 <= numRows <= 30)
Note:In "Yang Hui's Triangle", each number is the sum of the numbers on its upper left and upper right
Example one:
Import: numRows = 5 Output: [[1],[1,1],[1,2,1],[1,3,3,1],[ 1,4,6,4,1]]
Example 2:
Import: numRows = 1 Output: [[1]]
Idea:
We can change the direction of Yang Hui's triangle and change it to a ladder type, which may be more suitable for our understanding. For such ladder type data, we will naturally think of a two-dimensional array. So can Yang Hui's triangle be used as a two-dimensional array? What about array implementation?
For each row of data, the first and last ones are both 1, and the remaining data are the data at the same position in the previous row plus the previous data, which means we need to use it when implementing the data in each row. data from the previous row. In order to complete the above process more efficiently, we can use a two-dimensional sequence table instead of a two-dimensional array to operate.
After deciding to use a two-dimensional sequence table, we need to define a general sequence table to store the sequence table corresponding to each row, using the outer layerfor loop Determine how many rows there are, and use the inner layerfor loop to determine the elements of each row. The above determines the general framework of this question.
There are a few details to note:
Summary observation of Yang Hui's triangle, all1 positions are at the beginning and end of each line, the beginning is very simple and the inner layer “1” , we will add elements to the row sequence table (j == 0 || j == i) is equal, in summary, when the loop variable j variablefor loopand inner layer i for loop; is the starting position, and the For the last element, if we carefully observe its coordinates, we will find that its number of rows and columns are equal, which is the loop variable of the outer j=0 for loop
For data other than 1, we need to access the data in the previous row through .get( i-1) can get the sequence table of the previous row. After getting the sequence table of the previous row, get the j of this row respectively. to add this value to the sequence table of this row .add position elements, and then add them together. The result of the addition is the value we want, and finally Just use j-1 position and
Complete code:
class Solution {
public List<List<Integer>> generate(int numRows) {
List<List<Integer>> ret = new ArrayList<>();
for (int i = 0; i < numRows; i++) {
//新定义一行
List<Integer> row = new ArrayList<>();
for (int j = 0; j <= i; j++) {
if (j == 0 || j == i) {
row.add(1);
}else {
//添加上一行的数据（上一行的j-1和j位置）
row.add(ret.get(i-1).get(j-1) + ret.get(i-1).get(j));
}
}
//每一次将新的一行加入总的二维顺序表中
ret.add(row);
}
return ret;
}
}2. Merge two ordered arrays
Title: Give you two integer arrays arranged in non-decreasing order nums1 and nums2, and two integers m and n respectively represent . Please merge into so that the merged array is also Arrange in descending order . nums1nums2 nums2 nums1
Note:Ultimately, the merged array should not be returned by the function, but stored in the array nums1 . In order to cope with this situation, the initial length of nums1 is m + n, where the first m elements represent the elements that should be merged, and the last < /span> . is and should be ignored. The length of n elements are 0nums2n
Example one:
Import:nums1 = [1,2,3,0,0,0], m = 3, nums2 = [2,5,6], n = 3 Exit:[1,2,2,3,5,6] Solution:Demand sum [1,2,3] sum [2,5,6]. The result is [1,2,2, 3,5,6]
Example two:
Import:nums1 = [1], m = 1, nums2 = [], n = 0 Exit:[1] Solution:Demand sum [1] sum []. The result is [1]
Example three:
Input:nums1 = [0], m = 0, nums2 = [1], n = 1 Output: [1] Explanation: The arrays that need to be merged are [] and [1]. The combined result is [1]
Idea:
For an array, we want to merge it into a brand new array. We can use two pointers to point to the two arrays respectively, and then compare the two pointers with the smaller value, and put the smaller value into the new array, and pointer moves backward
There are a total of 4 situations for inserting data, and we will deal with them respectively:
nums1The data of is less than the data ofnums2: Put the data ofnums1into the new array, and after the pointer pA movenums1The data of is greater than the data ofnums2: Put the data ofnums2into the new array, and after the pointer pB movenums1The data of has been excluded, butnums2still has elements: put the elements ofnums2into the new array, and move the pointer pB back-
nums2The data of has been eliminated, butnums1still has elements: put the elements ofnums1into the new array, and move the pointer pA back
We use awhile loop to include all situation traversals, and then each judgment puts a piece of data into a new array through a temporary variable , in order to satisfy the question requirements, the final result is overwritten into array 1
Complete code:
class Solution {
public void merge(int[] nums1, int m, int[] nums2, int n) {
int pA = 0;
int pB = 0;
int cur = 0;
int[] sordNum = new int[m + n];
while (pA < m || pB < n) {
if (pA == m) {//nums1已满,放入nums2
cur = nums2[pB++];
} else if (pB == n) {//nums2已满，放入nums1
cur = nums1[pA++];
} else if (nums1[pA] < nums2[pB]) {//nums1的数据小于nums2
cur = nums1[pA++];
} else {//nums1的数据大于等于nums2
cur = nums2[pB++];
}
sordNum[pA + pB - 1] = cur;
}
for (int j = 0; j != m + n; ++j) {
nums1[j] = sordNum[j];
}
}
}3. Remove duplicates in an ordered array
Title: Give you an array non-strictly increasing arrangement nums , please delete repeated elements in situ so that each element appears only once . . Then return the number of unique elements in consistent of the elements should remain relative order , returns the new length of the array after deletion. The nums
Example one:
Import:nums = [1,1,2] Output:2, nums = [1,2,_]
Example two:
Import:nums = [0,0,1,1,1,2,2,3,3,4] Output:5, nums = [0,1,2,3,4]
Idea:
For thedeletion here, we can make clever use of it. Specifically, we can put the non-repeating elements in front of the array, and then We only return this part of non-repeating elements, thus achieving the goal of deletion in disguise
Complete code:
class Solution {
public int removeDuplicates(int[] nums) {
//快慢指针
int left = 0;
int right = 1;
while(right < nums.length){
if(nums[left] != nums[right]){
left++;
nums[left] = nums[right];
}
right++;
}
return left + 1;
}
} 
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