my_leetcode/java/src/com/iqiaoxu/code/Solution.java

package com.iqiaoxu.code;

import javafx.util.Pair;

import java.util.*;

public class Solution {
    public int xorOperation(int n, int start) {
        return (start & 3) < 2 ? ((n & 1) == 0) ? (n & 3) : (start + 2 * n - 3 + (n & 3)) : ((n & 1) == 0) ? ((start + (n - 1) * 2) ^ (start - 2 + (n & 3))) : (start + 1 - (n & 3));
    }

    public int mySqrt(int x) {
        double xHalf = 0.5 * x;
        long i = Double.doubleToLongBits(x);
        i = 0x1FF7A3BEA91D9B1BL + (i >> 1);
        double f = Double.longBitsToDouble(i);
        f = f * 0.5 + xHalf / f;
        f = f * 0.5 + xHalf / f;
        f = f * 0.5 + xHalf / f;
        return (int) f;
    }

    public int deleteAndEarn(int[] nums) {
        List<Integer> sum = new ArrayList<>(Collections.nCopies(Arrays.stream(nums).max().orElse(0) + 1, 0));
        Arrays.stream(nums).forEach(v -> sum.set(v, sum.get(v) + v));
        return sum.stream().reduce(new Pair<>(0, 0), ((acc, x) -> new Pair<>(acc.getValue(), Integer.max(acc.getKey() + x, acc.getValue()))), (a, b) -> new Pair<>(0, 0)).getValue();
    }

    public int rob(int[] nums) {
        return Arrays.stream(nums).boxed().reduce(new Pair<>(0, 0), ((a, x) -> new Pair<>(a.getValue(), Integer.max(a.getKey() + x, a.getValue()))), (a, b) -> new Pair<>(0, 0)).getValue();
    }

    public int rob2(int[] nums) {
        return Integer.max(Integer.max(Arrays.stream(nums).boxed().limit(nums.length - 1).reduce(new Pair<>(0, 0), ((a, x) -> new Pair<>(a.getValue(), Integer.max(a.getKey() + x, a.getValue()))), (a, b) -> new Pair<>(0, 0)).getValue(), Arrays.stream(nums).boxed().skip(1).reduce(new Pair<>(0, 0), ((a, x) -> new Pair<>(a.getValue(), Integer.max(a.getKey() + x, a.getValue()))), (a, b) -> new Pair<>(0, 0)).getValue()), nums[0]);
    }

    /**
     * 268.丢失的数字
     *
     * @param nums 数组
     * @return 丢失的数字
     * @see <a href="https://leetcode-cn.com/problems/missing-number/">原题链接</a>
     */
    public int missingNumber(int[] nums) {
        return new int[]{nums.length, 1, nums.length + 1, 0}[nums.length & 3] ^ Arrays.stream(nums).reduce(0, (o, n) -> o ^ n);
    }

    private boolean check(TreeNode l, TreeNode r) {
        return (l == null && r == null) || l != null && r != null && l.val == r.val && check(l.left, r.right) && check(l.right, r.left);
    }

    /**
     * 101.对称二叉树
     *
     * @param root 二叉树
     * @return 是否是对称二叉树
     * @see <a href="https://leetcode-cn.com/problems/symmetric-tree/">原题链接</a>
     */
    public boolean isSymmetric(TreeNode root) {
        /*if (root != null) {
            Queue<TreeNode> q = new LinkedList<>();
            q.offer(root.left);
            q.offer(root.right);
            while (!q.isEmpty()) {
                TreeNode l = q.poll();
                TreeNode r = q.poll();
                if (l == null && r == null) {
                    continue;
                }
                if (l == null || r == null || l.val != r.val) {
                    return false;
                }
                q.offer(l.left);
                q.offer(r.right);
                q.offer(l.right);
                q.offer(r.left);
            }
        }
        return true;*/
        /*return root == null || check(root.left, root.right);*/
        return root == null ||
               (root.left == null && root.right == null) ||
               !(root.left == null || root.right == null || root.left.val != root.right.val) &&
               isSymmetric(new TreeNode(0, root.left.left, root.right.right)) &&
               isSymmetric(new TreeNode(0, root.left.right, root.right.left));
    }

    /**
     * 640.求解方程
     * @param equation 方程式
     * @return 解
     * @see <a href="https://leetcode-cn.com/problems/solve-the-equation/">原题链接</a>
     */
    public String solveEquation(String equation) {
        int a = 0, b = 0, c = 0, sign = 1, left = 1;
        boolean hasC = false;
        for (byte z : equation.getBytes()) {
            switch (z) {
                case 'x':
                    a += hasC ? sign * left * c : sign * left;
                    c = 0;
                    hasC = false;
                    break;
                case '+':
                    b += sign * left * c;
                    c = 0;
                    sign = 1;
                    hasC = false;
                    break;
                case '-':
                    b += sign * left * c;
                    c = 0;
                    sign = -1;
                    hasC = false;
                    break;
                case '=':
                    b += sign * left * c;
                    c = 0;
                    sign = 1;
                    left = -1;
                    hasC = false;
                    break;
                default:
                    c = c * 10 + z - '0';
                    hasC = true;
                    break;
            }
        }
        b += sign * left * c;
        return a == 0 ? b == 0 ? "Infinite solutions" : "No solution" : "x=" + -b / a;
    }
}