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307. Range Sum Query - Mutable

2015-12-18 13:02 691 查看

题目:

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.

链接： http://leetcode.com/problems/range-sum-query-mutable/

题解：

这应该算是Range Query的经典题目之一了。也是通过这道题我第一次接触到了Segment Tree，也对Fenwick Tree有了一点了解。下面是用Segment Tree来做的。 Segment Tree线段树每一个节点都是一段线段，有start和end，然后还可以有其他的值，比如区间和sum，区间最大值max，区间最小值min。我们可以用自底向上构建二叉树的方式构建Segment Tree，这个过程也有点类似于Bottom-up的merge sort，思想也是Divide and Conquer。完毕之后就可以在O(logn)的时间update，或者得到range Sum。其实更好的方法是使用Fenwick Tree， Fenwick Tree(Binary Indexed Tree)在处理Range Query真的是一绝，构造简练，原理也精妙，还可以扩展到多维，一定要好好学一学。

Time Complexity - O(n) build， O(logn) update, O(logn) rangeSum， Space Complexity - O(n)

public class NumArray {
private class SegmentTreeNode {
public int start;
public int end;
public int sum;
public SegmentTreeNode left, right;
public SegmentTreeNode(int start, int end) {
this.start = start;
this.end = end;
this.sum = 0;
}
}

private SegmentTreeNode root;

public NumArray(int[] nums) {
this.root = buildTree(nums, 0, nums.length - 1);
}

public void update(int i, int val) {
update(root, i, val);
}

private void update(SegmentTreeNode node, int position, int val) {
if(node.start == position && node.end == position) {
node.sum = val;
return;
}
int mid = node.start + (node.end - node.start) / 2;
if(position <= mid) {
update(node.left, position, val);
} else {
update(node.right, position, val);
}
node.sum = node.left.sum + node.right.sum;
}

public int sumRange(int i, int j) {
return sumRange(root, i, j);
}

private int sumRange(SegmentTreeNode node, int lo, int hi) {
if(node.start == lo && node.end == hi) {
return node.sum;
}
int mid = node.start + (node.end - node.start) / 2;
if(hi <= mid) {
return sumRange(node.left, lo, hi);
} else if (lo > mid) {
return sumRange(node.right, lo, hi);
} else {
return sumRange(node.left, lo, mid) + sumRange(node.right, mid + 1, hi);
}
}

private SegmentTreeNode buildTree(int[] nums, int lo, int hi) {
if(lo > hi) {
return null;
} else {
SegmentTreeNode node = new SegmentTreeNode(lo, hi);
if(lo == hi) {
node.sum = nums[lo];
} else {
int mid = lo + (hi - lo) / 2;
node.left = buildTree(nums, lo, mid);
node.right = buildTree(nums, mid + 1, hi);
node.sum = node.left.sum + node.right.sum;
}
return node;
}
}
}

// 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);

Fenwick Tree: (Binary Indexed Tree) (树状数组)

很有意思的构建，以数组nums = {1, 2, 3, 4, 5, 6, 7, 8}为例，这个数组长度为8。跟dynamic programming的预处理很像，我们先建立一个长度为nums.length + 1 = 9的数组BIT。接下来遍历数组nums，对BIT数组进行update(i + 1, nums[i])。这里BIT数组每个值BIT[i]代表nums数组里在i之前的部分元素和。原理是像自然数可以被表示为2n的和一样，把nums数组里到0到i的sum表示成2n的和，从而导致update和rangeSum都可以用O(logn)的时间求出来。这里构建的时候可以有几种写法，主要就是利用当前i的least significante 1来确定到底BIT[i]要保存多少原数组的值。这里借用algorithmist的原话"Every index in the cumulative sum array, say i, is responsible for the cumulative sum from the index i to (i - (1<<r) + 1)。" 构建过程中可以用 (i & -i)来找到least significate 1，之后来进行i = i + (i & -i)来尝试从小到大计算下一个BIT数组中被影响的元素。而rangeSum的时候则使用i = i - (i & -i)来从大到小查找从0到i - 1的sum。

构建过程 - update，给定数组nums = {1，2, 3, 4, 5, 6, 7, 8}

BIT[0] = 0

BIT[1] = nums[0] = 1 = 1

BIT[2] = nums[0] + nums[1] = 1 + 2 = 3

BIT[3] = nums[2] = 3 = 3

BIT[4] = nums[0] + nums[1] + nums[2] + nums[3] = 1+ 2 + 3 + 4 = 10

BIT[5] = nums[4] = 5 = 5

BIT[6] = nums[4] + nums[5] = 5 + 6 = 11

BIT[7] = nums[6] = 7 = 7

BIT[8] = nums[0] + nums[1] + nums[2] + nums[3] + nums[4] + nums[5] + nums[6] + nums[7] = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 36

求Sum过程，通过 sum = BIT[i + 1]; i = i - (i & -i); 从大到小迭代来计算。

sum(0) = BIT[1]

sum(1) = BIT[2]

sum(2) = BIT[3] + BIT[2]

sum(3) = BIT[4]

sum(4) = BIT[5] + BIT[4]

sum(5) = BIT[6] + BIT[4]

sum(6) = BIT[7] + BIT[6] + BIT[4]

sum(7) = BIT[8]

得到sum(i)以后就可以相减来计算range sum了。

Time Complexity - O(nlogn) build， O(logn) update， O(logn) rangeSum， Space Complexity - O(n)

public class NumArray {
private int BIT[];               // Binary Indexed Tree = Fenwick Tree
private int[] nums;

public NumArray(int[] nums) {
BIT = new int[nums.length + 1];
for(int i = 0; i < nums.length; i++) {
init(i + 1, nums[i]);
}
this.nums = nums;
}

private void init(int i, int val) {
while(i < BIT.length) {
BIT[i] += val;
i = i + (i & -i);
}
}

public void update(int i, int val) {
int delta = val - nums[i];
nums[i] = val;
init(i + 1, delta);
}

public int sumRange(int i, int j) {
return getSum(j + 1) - getSum(i);
}

private int getSum(int i) {
int sum = 0;
while(i > 0) {
sum += BIT[i];
i = i - (i & -i);
}
return sum;
}
}

题外话：今天去NYU图书馆自习，正值考期，人山人海的。我带了电脑却忘记带插座，无奈只能用Java 在线的编译器https://www.compilejava.net/，不过这个真的还挺好用，除了不能debug，其他都可以，nice。晚上吃了Hakata Tonton，4个人大概人均$50+，并没有想象的那么好吃，以后还是要攒机会去日本玩。刷题群里大家对Heapify有了热烈的讨论，我自己认为Heapify主要有两种，Bottom-up (swim)和Top-down(sink)。也要复习一下Priority Queue的implementation。要多看Sedgewick的课件和sample code才行。在GitHub发现有个人叫indy256，实现了好多好多高级数据结构，有2d-fenwick tree，以及O(n)的Suffix Tree。大牛和普通人的距离真的好遥远，我还是继续努力。

二刷:

暂时只用了segment tree。 Fenwick Tree以后再理解。

Java:

Segment Tree:

public class NumArray {
private SegmentTreeNode root;
private int[] nums;

public NumArray(int[] nums) {
this.nums = nums;
this.root = buildTree(0, nums.length - 1);
}

void update(int i, int val) {
update(root, i, val);
}

private void update(SegmentTreeNode node, int pos, int val) {
if (node == null) return;
if (node.start == pos && node.end == pos) {
node.val = val;
nums[pos] = val;
return;
}
int mid = node.start + (node.end - node.start) / 2;
if (pos <= mid) {
update(node.left, pos, val);
} else {
update(node.right, pos, val);
}
node.val = node.left.val + node.right.val;
}

public int sumRange(int i, int j) {
return sumRange(root, i, j);
}

private int sumRange(SegmentTreeNode node, int lo, int hi) {
if (lo > hi) return 0;
if (node.start == lo && node.end == hi) return node.val;
int mid = node.start + (node.end - node.start) / 2;
if (hi <= mid) {
return sumRange(node.left, lo, hi);
} else if (lo > mid) {
return sumRange(node.right, lo, hi);
} else {
return sumRange(node.left, lo, mid) + sumRange(node.right, mid + 1, hi);
}
}

private SegmentTreeNode buildTree(int lo, int hi) {
if (lo > hi) return null;
SegmentTreeNode node = new SegmentTreeNode(lo, hi);
if (lo == hi) {
node.val = nums[lo];
} else {
int mid = lo + (hi - lo) / 2;
node.left = buildTree(lo, mid);
node.right = buildTree(mid + 1, hi);
node.val = node.left.val + node.right.val;
}
return node;
}

private class SegmentTreeNode {
int start;
int end;
int val;
SegmentTreeNode left, right;

public SegmentTreeNode(int start, int end) {
this.start = start;
this.end = end;
this.val = 0;
}
}
}

// 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);

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