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package class35;
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import java.util.ArrayList;
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import java.util.TreeMap;
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public class Problem_0673_NumberOfLongestIncreasingSubsequence {
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// 好理解的方法，时间复杂度O(N^2)
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public static int findNumberOfLIS1(int[] nums) {
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if (nums == null || nums.length == 0) {
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return 0;
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}
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int n = nums.length;
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int[] lens = new int[n];
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int[] cnts = new int[n];
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lens[0] = 1;
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cnts[0] = 1;
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int maxLen = 1;
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int allCnt = 1;
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for (int i = 1; i < n; i++) {
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int preLen = 0;
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int preCnt = 1;
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for (int j = 0; j < i; j++) {
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if (nums[j] >= nums[i] || preLen > lens[j]) {
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continue;
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}
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if (preLen < lens[j]) {
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preLen = lens[j];
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preCnt = cnts[j];
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} else {
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preCnt += cnts[j];
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}
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}
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lens[i] = preLen + 1;
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cnts[i] = preCnt;
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if (maxLen < lens[i]) {
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maxLen = lens[i];
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allCnt = cnts[i];
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} else if (maxLen == lens[i]) {
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allCnt += cnts[i];
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}
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}
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return allCnt;
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}
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// 优化后的最优解，时间复杂度O(N*logN)
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public static int findNumberOfLIS2(int[] nums) {
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if (nums == null || nums.length == 0) {
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return 0;
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}
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ArrayList<TreeMap<Integer, Integer>> dp = new ArrayList<>();
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int len = 0;
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int cnt = 0;
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for (int num : nums) {
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// num之前的长度，num到哪个长度len+1
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len = search(dp, num);
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// cnt : 最终要去加底下的记录，才是应该填入的value
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if (len == 0) {
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cnt = 1;
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} else {
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TreeMap<Integer, Integer> p = dp.get(len - 1);
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cnt = p.firstEntry().getValue() - (p.ceilingKey(num) != null ? p.get(p.ceilingKey(num)) : 0);
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}
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if (len == dp.size()) {
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dp.add(new TreeMap<Integer, Integer>());
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dp.get(len).put(num, cnt);
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} else {
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dp.get(len).put(num, dp.get(len).firstEntry().getValue() + cnt);
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}
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}
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return dp.get(dp.size() - 1).firstEntry().getValue();
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}
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// 二分查找，返回>=num最左的位置
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public static int search(ArrayList<TreeMap<Integer, Integer>> dp, int num) {
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int l = 0, r = dp.size() - 1, m = 0;
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int ans = dp.size();
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while (l <= r) {
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m = (l + r) / 2;
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if (dp.get(m).firstKey() >= num) {
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ans = m;
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r = m - 1;
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} else {
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l = m + 1;
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}
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}
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return ans;
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}
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}
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