leetcode 307. Range Sum Query

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.

题意很简单，但是我的做法总是超时，想了好久没想到更好的方法，网上看了个答案，原来还可以这么做。树状数组没学过，也不知道又该怎么做，所以这道题就先记着吧。

要和这道题leetcode 304. Range Sum Query 2D - Immutable 子矩阵求和 + DP 和 这道题leetcode 303. Range Sum Query - Immutable 字串求和 + DP 一起学习。

代码如下：

import java.util.Arrays;/* *使用树状数组这一个数据结构，自行掌握吧， *http://www.cnblogs.com/xudong-bupt/p/3484080.html *或者使用线段树 * *还是先记着吧，第一次接触 * * */public class NumArray {    /**     * 使用     * Binary Indexed Trees (BIT or Fenwick tree):     * https://www.topcoder.com/community/data-science/data-science-     * tutorials/binary-indexed-trees/     *      * Example: given an array a[0]...a[7], we use a array BIT[9] to     * represent a tree, where index [2] is the parent of [1] and [3], [6]     * is the parent of [5] and [7], [4] is the parent of [2] and [6], and     * [8] is the parent of [4]. I.e.,     *      * BIT[] as a binary tree:     *            ______________*     *            ______*     *            __*     __*     *            *   *   *   *     * indices: 0 1 2 3 4 5 6 7 8     *      * BIT[i] = ([i] is a left child) ? the partial sum from its left most     * descendant to itself : the partial sum from its parent (exclusive) to     * itself. (check the range of "__").     *      * Eg. BIT[1]=a[0], BIT[2]=a[1]+BIT[1]=a[1]+a[0], BIT[3]=a[2],     * BIT[4]=a[3]+BIT[3]+BIT[2]=a[3]+a[2]+a[1]+a[0],     * BIT[6]=a[5]+BIT[5]=a[5]+a[4],     * BIT[8]=a[7]+BIT[7]+BIT[6]+BIT[4]=a[7]+a[6]+...+a[0], ...     *      * Thus, to update a[1]=BIT[2], we shall update BIT[2], BIT[4], BIT[8],     * i.e., for current [i], the next update [j] is j=i+(i&-i) //double the     * last 1-bit from [i].     *      * Similarly, to get the partial sum up to a[6]=BIT[7], we shall get the     * sum of BIT[7], BIT[6], BIT[4], i.e., for current [i], the next     * summand [j] is j=i-(i&-i) // delete the last 1-bit from [i].     *      * To obtain the original value of a[7] (corresponding to index [8] of     * BIT), we have to subtract BIT[7], BIT[6], BIT[4] from BIT[8], i.e.,     * starting from [idx-1], for current [i], the next subtrahend [j] is     * j=i-(i&-i), up to j==idx-(idx&-idx) exclusive. (However, a quicker     * way but using extra space is to store the original array.)     */    int[] nums=null;    int[] BIT=null;    public NumArray(int[] nums)     {        this.nums=new int[nums.length];        BIT=new int[nums.length+1];        Arrays.fill(BIT, 0);        for(int i=0;i<nums.length;i++)        {            this.nums[i]=nums[i];            init(i, nums[i]);        }    }    public int lowBit(int i)     {        return i&(-i);    }    public void init(int i, int val)     {        i=i+1;        while(i<=nums.length)        {            BIT[i]+=val;            i+=lowBit(i);        }    }    void update(int i, int val)    {        int diff=val-nums[i];        nums[i]=val;        init(i, diff);    }    public int getSum(int i)     {        i=i+1;        int sum=0;        while(i>0)        {            sum+=BIT[i];            i-=lowBit(i);        }        return sum;    }    public int sumRange(int i, int j)     {        return getSum(j)-getSum(i-1);    }}// Your NumArray object will be instantiated and called as such:// NumArray numArray = new NumArray(nums);// numArray.sumRange(0, 1);// numArray.update(1, 10);// numArray.sumRange(1, 2);/*public class NumArray{    public static int[] num=null;    public static int[] sum=null;    public NumArray(int[] nums)    {        if(nums==null || nums.length<=0)            return ;        num=new int[nums.length];        sum=new int[nums.length];        sum[0]=nums[0];        num[0]=nums[0];        for(int i=1;i<nums.length;i++)        {             sum[i]=sum[i-1]+nums[i];             num[i]=nums[i];        }    }    public void update(int i, int val)     {        if(i<0 || i>=num.length)            return ;        num[i]=val;        if(i==0)        {             sum[0]=val;             for(int j=1;j<num.length;j++)                sum[j]=sum[j-1]+num[j];        }else         {            for(int j=i;j<num.length;j++)                sum[j]=sum[j-1]+num[j];        }    }    public int sumRange(int i, int j)    {        if(sum==null || sum.length<=0 || i<0 || i>=num.length || j<0 || j>=num.length)            return -1;        if(i==0)            return sum[j];        else             return sum[j]-sum[i-1];    }}*/