标签：
Given an integer array (index from 0 to n-1, where n is the size of this array), and an query list. Each query has two integers [start, end]. For each query, calculate the sum number between index start and end in the given array, return the result list.
For array [1,2,7,8,5], and queries [(0,4),(1,2),(2,4)], return [23,9,20].
Analysis:
Use an array to save the sum from 0 to i. Then for query [i, j], we shoud return sum[j] - sum[i - 1].
1 /** 2 * Definition of Interval: 3 * public classs Interval { 4 * int start, end; 5 * Interval(int start, int end) { 6 * this.start = start; 7 * this.end = end; 8 * } 9 */ 10 public class Solution { 11 /** 12 *@param A, queries: Given an integer array and an query list 13 *@return: The result list 14 */ 15 public ArrayList<Long> intervalSum(int[] A, 16 ArrayList<Interval> queries) { 17 18 ArrayList<Long> list = new ArrayList<Long>(); 19 if (A == null || A.length == 0) return list; 20 if (queries == null || queries.size() == 0) return list; 21 22 long[] sum = new long[A.length]; 23 24 for (int i = 0; i < sum.length; i++) { 25 if (i == 0) { 26 sum[i] = A[i]; 27 } else { 28 sum[i] += sum[i - 1] + A[i]; 29 } 30 } 31 32 for (int i = 0; i < queries.size(); i++) { 33 Interval interval = queries.get(i); 34 if (interval.start == 0) { 35 list.add(sum[interval.end]); 36 } else { 37 list.add(sum[interval.end] - sum[interval.start - 1]); 38 } 39 } 40 return list; 41 } 42 }
Given an integer array in the construct method, implement two methods query(start, end)and modify(index, value):
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.Analysis:
As the value in the array may change, so it is better to build a segement tree. If the value in the tree is changed, we also need to update its parent node.
1 public class Solution { 2 /* you may need to use some attributes here */ 3 4 SegmentTreeNode root; 5 6 /** 7 * @param A: 8 * An integer array 9 */ 10 public Solution(int[] A) { 11 if (A == null || A.length == 0) 12 return; 13 root = build(A, 0, A.length - 1); 14 } 15 16 private SegmentTreeNode build(int[] A, int start, int end) { 17 if (A == null || start < 0 || end >= A.length) 18 return null; 19 SegmentTreeNode root = new SegmentTreeNode(start, end, 0); 20 if (start == end) { 21 root.sum = A[start]; 22 } else { 23 int mid = (end - start) / 2 + start; 24 root.left = build(A, start, mid); 25 root.right = build(A, mid + 1, end); 26 root.sum = root.left.sum + root.right.sum; 27 } 28 return root; 29 } 30 31 public long query(int start, int end) { 32 if (root == null || start > end) 33 return 0; 34 return helper(root, Math.max(0, start), Math.min(end, root.end)); 35 } 36 37 public long helper(SegmentTreeNode root, int start, int end) { 38 if (root.start == start && root.end == end) { 39 return root.sum; 40 } 41 42 int mid = (root.start + root.end) / 2; 43 if (start >= mid + 1) { 44 return helper(root.right, start, end); 45 } else if (end <= mid) { 46 return helper(root.left, start, end); 47 } else { 48 return helper(root.left, start, mid) + helper(root.right, mid + 1, end); 49 } 50 } 51 52 public void modify(int index, int value) { 53 if (root == null) 54 return; 55 if (index < root.start && index > root.end) 56 return; 57 modifyHelper(root, index, value); 58 } 59 60 public void modifyHelper(SegmentTreeNode root, int index, int value) { 61 if (root.start == root.end && root.start == index) { 62 root.sum = value; 63 return; 64 } 65 66 int mid = (root.start + root.end) / 2; 67 if (index >= mid + 1) { 68 modifyHelper(root.right, index, value); 69 } else { 70 modifyHelper(root.left, index, value); 71 } 72 root.sum = root.left.sum + root.right.sum; 73 74 } 75 } 76 77 class SegmentTreeNode { 78 public int start, end, sum; 79 public SegmentTreeNode left, right; 80 81 public SegmentTreeNode(int start, int end, int sum) { 82 this.start = start; 83 this.end = end; 84 this.sum = sum; 85 this.left = this.right = null; 86 } 87 }
标签：
原文地址：http://www.cnblogs.com/beiyeqingteng/p/5668864.html
