[LeetCode]493. Reverse Pairs

来源：互联网发布：安卓导航软件编辑：程序博客网时间：2024/06/05 01:02

https://leetcode.com/problems/reverse-pairs/?tab=Description

找出满足i < j && nums[i] > 2 * nums[j]的所有对数

树状数组：

public class Solution {    public int reversePairs(int[] nums) {        int[] copy = Arrays.copyOf(nums, nums.length);        int[] bit = new int[nums.length + 1];        Arrays.sort(copy);        int res = 0;        for (int num : nums) {            res += search(bit, index(copy, 2L * num + 1));            insert(bit, index(copy, num));        }        return res;    }    // 找比当前值大的，要往后找    private int search(int[] bit, int i) {        int sum = 0;        while (i < bit.length) {            sum += bit[i];            i += (i & -i);        }        return sum;    }    // 更新所有小于等于位置i的值所在的bit    private void insert(int[] bit, int i) {        while (i > 0) {            bit[i]++;            i -= (i & -i);        }    }    private int index(int[] nums, long val) {        int i = 0;        int j = nums.length - 1;        while (i <= j) {            int mid = i + (j - i) / 2;            if (nums[mid] >= val) {                j = mid - 1;            } else {                i = mid + 1;            }        }        // 返回bit内的index，要加一        return i + 1;    }}

BST，在[0, 1, 2, 3.....]大数据量时会超时，主要看思路

public class Solution {    public int reversePairs(int[] nums) {        int cnt = 0;        TreeNode root = null;        for (int num : nums) {            cnt += search(root, 2L * num + 1);            root = insert(num, root);        }        return cnt;    }    private TreeNode insert(int val, TreeNode root) {        if (root == null) {            return new TreeNode(val);        } else if (val == root.val) {            root.cnt++;        } else if (root.val > val) {            root.left = insert(val, root.left);        } else {            root.cnt++;            root.right = insert(val, root.right);        }        return root;    }    private int search(TreeNode root, long val) {        if (root == null) {            return 0;        } else if (root.val == val) {            return root.cnt;        } else if (root.val > val) {            return root.cnt + search(root.left, val);        } else {            return search(root.right, val);        }    }}class TreeNode {    int val;    int cnt;    TreeNode left;    TreeNode right;    public TreeNode(int val) {        this.val = val;        this.cnt = 1;    }}

归并排序，合并过程中从左右两个子数组中找到满足条件的对数，因为i < j始终满足，只要找nums[i] > 2 * nums[j]即可

public class Solution {    public int reversePairs(int[] nums) {        return reversePairs(nums, 0, nums.length - 1);    }    private int reversePairs(int[] nums, int beg, int end) {        if (beg >= end) {            return 0;        }        int l = beg;        int m = beg + (end - beg) / 2;        int r = m + 1;        int p = m + 1;        int[] merge = new int[end - beg + 1];        int k = 0;        int res = reversePairs(nums, beg, m) + reversePairs(nums, m + 1, end);        while (l <= m) {            while (p <= end && nums[l] > 2L * nums[p]) {                p++;            }            res += p - (m + 1);            while (r <= end && nums[l] >= nums[r]) {                merge[k++] = nums[r++];            }            merge[k++] = nums[l++];        }        while (r <= end) {            merge[k++] = nums[r++];        }        System.arraycopy(merge, 0, nums, beg, merge.length);        return res;    }}

0 0