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  • Lintcode: Interval Sum II

    Given an integer array in the construct method, implement two methods query(start, end) and modify(index, value):
    
    For query(start, end), return the sum from index start to index end in the given array.
    For modify(index, value), modify the number in the given index to value
    Have you met this question in a real interview? Yes
    Example
    Given array A = [1,2,7,8,5].
    
    query(0, 2), return 10.
    modify(0, 4), change A[0] from 1 to 4.
    query(0, 1), return 6.
    modify(2, 1), change A[2] from 7 to 1.
    query(2, 4), return 14.
    Note
    We suggest you finish problem Segment Tree Build, Segment Tree Query and Segment Tree Modify first.
    
    Challenge
    O(logN) time for query and modify.

    Segment Tree:

     1 public class Solution {
     2     /* you may need to use some attributes here */
     3     class SegmentTreeNode {
     4         long sum;
     5         int start;
     6         int end;
     7         SegmentTreeNode left;
     8         SegmentTreeNode right;
     9         SegmentTreeNode(int start, int end) {
    10             this.sum = 0;
    11             this.start = start;
    12             this.end = end;
    13             this.left = null;
    14             this.right = null;
    15         }
    16     }
    17     
    18     SegmentTreeNode root;
    19 
    20     /**
    21      * @param A: An integer array
    22      */
    23     public Solution(int[] A) {
    24         // write your code here
    25         if (A == null || A.length==0) return;
    26         root = build(A, 0, A.length-1);
    27         
    28     }
    29     
    30     public SegmentTreeNode build(int[] A, int start, int end) {
    31         SegmentTreeNode cur = new SegmentTreeNode(start, end);
    32         if (start == end) cur.sum = A[start];
    33         else {
    34             int mid = (start + end)/2;
    35             cur.left = build(A, start, mid);
    36             cur.right = build(A, mid+1, end);
    37             cur.sum = cur.left.sum + cur.right.sum;
    38         }
    39         return cur;
    40     }
    41     
    42     /**
    43      * @param start, end: Indices
    44      * @return: The sum from start to end
    45      */
    46     public long query(int start, int end) {
    47         // write your code here
    48         return queryTree(root, start, end);
    49     }
    50     
    51     public long queryTree(SegmentTreeNode cur, int start, int end) {
    52         if (cur.start==start && cur.end==end) return cur.sum;
    53         int mid = (cur.start + cur.end)/2;
    54         if (end <= mid) return queryTree(cur.left, start, end);
    55         else if (start > mid) return queryTree(cur.right, start, end);
    56         else return queryTree(cur.left, start, mid) + queryTree(cur.right, mid+1, end);
    57     }
    58     
    59     /**
    60      * @param index, value: modify A[index] to value.
    61      */
    62     public void modify(int index, int value) {
    63         // write your code here
    64         modifyTree(root, index, value);
    65     }
    66     
    67     public void modifyTree(SegmentTreeNode cur, int index, int val) {
    68         if (cur.start == cur.end) {
    69             cur.sum = val;
    70             return;
    71         }
    72         int mid = (cur.start + cur.end)/2;
    73         if (index <= mid) modifyTree(cur.left, index, val);
    74         else modifyTree(cur.right, index, val);
    75         cur.sum = cur.left.sum + cur.right.sum;
    76     }
    77 }
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  • 原文地址:https://www.cnblogs.com/EdwardLiu/p/5176677.html
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