博客记录-day160-力扣

一、力扣

1、排序奇升偶降链表

排序奇升偶降链表

思路：可以直接使用归并排序，也可以先拆分为两个链表，再对第二个反转，最后再合并。

public class Solution {
    public ListNode sortLinkedList (ListNode head) {
        // write code here
        ListNode fir = new ListNode(0);
        ListNode firlas = fir;
        ListNode sec = new ListNode(0);
        ListNode seclas = sec;
        int count = 0;
        while (head != null) {
            if (count % 2 == 0) {
                firlas.next = head;
                firlas = head;
                head = head.next;
                firlas.next = null;
            } else {
                seclas.next = head;
                seclas = head;
                head = head.next;
                seclas.next = null;
            }
            count++;
        }
        ListNode newSec = reverse(sec.next);

        return merge(newSec, fir.next);
    }
    public ListNode reverse(ListNode head) {
        ListNode pre = null;
        while (head != null) {
            ListNode headNext = head.next;
            head.next = pre;
            pre = head;
            head = headNext;
        }
        return pre;
    }
    public ListNode merge(ListNode head1, ListNode head2) {
        ListNode dummy = new ListNode(0);
        ListNode last = dummy;
        while (head1 != null && head2 != null) {
            if (head1.val > head2.val) {
                last.next = head2;
                last = head2;
                head2 = head2.next;
                last.next = null;
            } else {
                last.next = head1;
                last = head1;
                head1 = head1.next;
                last.next = null;
            }
        }
        last.next = head1 == null ? head2 : head1;
        return dummy.next;
    }
}

2、零钱兑换 II

518. 零钱兑换 II

完全背包是从i+1，01背包是从i。

class Solution {
    public int change(int amount, int[] coins) {
        int n = coins.length;
        int[][] dp = new int[n + 1][amount + 1];
        dp[0][0] = 1;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j <= amount; j++) {
                if (j < coins[i]) {
                    dp[i + 1][j] = dp[i][j];
                } else {
                    dp[i + 1][j] = dp[i][j] + dp[i + 1][j - coins[i]];
                }
            }
        }
        return dp[n][amount];
    }
}

3、Pow(x, n)

50. Pow(x, n)

class Solution {
    public double myPow(double x, int n) {
        double res = 1;
        long ans = n;
        if (ans < 0) {
            ans = -ans;
            x = 1 / x;
        }
        while (n != 0) {
            if ((n & 1) == 1) {
                res *= x;
            }
            x = x * x;
            n = n / 2;
        }
        return res;
    }
}

4、判断子序列

392. 判断子序列

class Solution {
    public boolean isSubsequence(String s, String t) {
        int m = s.length(), n = t.length();
        int[][] dp = new int[m + 1][n + 1];
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (s.charAt(i) == t.charAt(j)) {
                    dp[i + 1][j + 1] = dp[i][j] + 1;
                } else {
                    dp[i + 1][j + 1] = Math.max(dp[i + 1][j], dp[i][j + 1]);
                }
            }
        }
        return dp[m][n] == m;
    }
}