Lint 202.Segment Tree Query

二刷
一刷的思路应该是错的
正确的做法是每次不改变传入的start和end
对每次的range分Partial Overlap, Total Overlap和No Overlap三种情况讨论

public class Solution {
    /**
     *@param root, start, end: The root of segment tree and
     *                         an segment / interval
     *@return: The maximum number in the interval [start, end]
     */
    public int query(SegmentTreeNode root, int start, int end) {
        // write your code here
        int left = root.start;
        int right = root.end;
        //1.No Overlap
        if (root == null || start > end || end < left || start > right) return Integer.MIN_VALUE;
        //2.Total Overlap
        if (start <= left && end >= right) return root.max;
        //3.Partial Overlap
        return Math.max(query(root.left, start, end), query(root.right, start, end));
    }
}


一刷

写了一个bug
 int mid = (root.start + root.end) / 2; 此处mid应该是树Node的mid
而不应该是query的mid
Time O(logn)

/**
 * Definition of SegmentTreeNode:
 * public class SegmentTreeNode {
 *     public int start, end, max;
 *     public SegmentTreeNode left, right;
 *     public SegmentTreeNode(int start, int end, int max) {
 *         this.start = start;
 *         this.end = end;
 *         this.max = max
 *         this.left = this.right = null;
 *     }
 * }
 */
public class Solution {
    /**
     *@param root, start, end: The root of segment tree and
     *                         an segment / interval
     *@return: The maximum number in the interval [start, end]
     */
    public int query(SegmentTreeNode root, int start, int end) {
        // write your code here
        if (start > end) throw new IllegalArgumentException();
        if (root.start == start && root.end == end) return root.max;
        int mid = (root.start + root.end) / 2;
        if (end <= mid) return query(root.left, start, end);
        else if (start > mid) return query(root.right, start, end);
        else return Math.max(query(root.left, start, mid), query(root.right, mid + 1, end));
    }
}