"Leetcode" Russian Doll envelopes

Title Link: https://leetcode.com/problems/russian-doll-envelopes/

Topic:

You had a number of envelopes with widths and heights given as a pair of integers (w, h) . One envelope can fit into another if and only if both the width and height of one envelope is greater than the width and h Eight of the other envelope.

What is the maximum number of envelopes can Russian doll? (Put one inside other)

Example:
Given envelopes = [[5,4],[6,4],[6,7],[2,3]] , the maximum number of envelopes you can Russian doll are 3 ([2,3] = [5,4] = [6,7]).

Ideas:

Similar to the longest growth substring, the difference is that the conditions for determining growth are different, and the array needs to be sorted first.

C[i] represents a size from 0 to I envelopes, the maximum number that can be contained, and the maximum number of envelopes that is recorded to I with a array.

State transition equation: c[i]=max{c[j]+1} envelopes[i][0]>envelopes[j][0]&&envelopes[i][1]> ENVELOPES[J][1] 0<=j<i

Sorting time complexity O (NLOGN), DP time complexity O (n^2), DP when in fact can be further using binary search optimization to O (NLOGN). However, because the following practice can also be too lazy to optimize the AC. = =

Algorithm:

private void QuickSort (int[][] array, int beg, int end) {if (Beg >= end | | array = NULL) Return;int p = partition (array , beg, end), QuickSort (array, Beg, p-1), QuickSort (array, p + 1, end);} private int partition (int[][] array, int beg, int end) {int first = Array[beg][0];int i = Beg, j = End;while (I < j) {W Hile (Array[i][0] <= First && i < end) {i++;} while (Array[j][0] > First && J >= Beg) {j--;} if (I < j) {int tmp[] = array[i];array[i] = array[j];array[j] = tmp;}} if (j! = Beg) {int tmp[] = array[j];array[j] = Array[beg];array[beg] = tmp;} return J;} public int maxenvelopes (int[][] envelopes) {if (envelopes.length = = 0) {return 0;} int a[][] = new int[envelopes.length][2];//to I last set of baby's size int c[] = new int[envelopes.length];//C[i] represents from 0 to I envelopes, Maximum number of possible c[0] = 1;quicksort (envelopes, 0, envelopes.length-1); a[0] = Envelopes[0];int max = 1;for (int i = 1; i < env Elopes.length; i++) {int m = 1;boolean flag = true;for (int j = i-1; J >= 0; j--) {if (Envelopes[i][0] > A[j][0] && envelopes[i][1] > a[j][1]) {flag = false;if (M <= c[j] + 1) {m = C[j] + 1;a[i] = Envelopes[i];}} if (flag) {A[i] = envelopes[i];} C[i] = M;max = Math.max (max, m);} return Max;}

"Leetcode" Russian Doll envelopes