Given an integer array nums, find the sum of the elements between indices i and j (i ≤ j), inclusive.
The update(i, val) function modifies nums by updating the element at index i to val.
Example:
Given nums = [1, 3, 5] sumRange(0, 2) -> 9 update(1, 2) sumRange(0, 2) -> 8
Note:
- The array is only modifiable by the update function.
- You may assume the number of calls to update and sumRange function is distributed evenly.
1 public class NumArray { 2 private int[] tree; 3 private int[] num; 4 5 public NumArray(int[] nums) { 6 tree = new int[nums.length]; 7 num = new int[nums.length]; 8 9 Arrays.fill(tree,0); 10 Arrays.fill(num,0); 11 for (int i=0;i<nums.length;i++){ 12 update(i,nums[i]); 13 } 14 } 15 16 void update(int i, int val) { 17 int delta = val-num[i]; 18 num[i] = val; 19 int ind = i+1; 20 21 while (ind<=tree.length){ 22 tree[ind-1] += delta; 23 ind += ind & (-ind); 24 } 25 26 } 27 28 private int sum(int i){ 29 int ind = i+1; 30 int sum = 0; 31 32 while (ind>0){ 33 sum += tree[ind-1]; 34 ind -= ind & (-ind); 35 } 36 return sum; 37 } 38 39 public int sumRange(int i, int j) { 40 int a = sum(j); 41 int b = sum(i-1); 42 return a-b; 43 } 44 45 46 } 47 48 49 // Your NumArray object will be instantiated and called as such: 50 // NumArray numArray = new NumArray(nums); 51 // numArray.sumRange(0, 1); 52 // numArray.update(1, 10); 53 // numArray.sumRange(1, 2);
Solution: it is an entry-level problem about Binary Indexed Tree.
Check BIT out: http://www.geeksforgeeks.org/binary-indexed-tree-or-fenwick-tree-2/