题目
给你一个数组 nums ,请你完成两类查询。
其中一类查询要求 更新 数组 nums 下标对应的值
另一类查询要求返回数组 nums 中索引 left 和索引 right 之间( 包含 )的nums元素的 和 ,其中 left <= right
实现 NumArray 类:NumArray(int[] nums) 用整数数组 nums 初始化对象
void update(int index, int val) 将 nums[index] 的值 更新 为 val
int sumRange(int left, int right) 返回数组 nums 中索引 left 和索引 right 之间( 包含 )的nums元素的 和 (即,nums[left] + nums[left + 1], …, nums[right])示例 1:
输入:
[“NumArray”, “sumRange”, “update”, “sumRange”]
[[[1, 3, 5]], [0, 2], [1, 2], [0, 2]]
输出:
[null, 9, null, 8]解释:
NumArray numArray = new NumArray([1, 3, 5]);
numArray.sumRange(0, 2); // 返回 1 + 3 + 5 = 9
numArray.update(1, 2); // nums = [1,2,5]
numArray.sumRange(0, 2); // 返回 1 + 2 + 5 = 8提示:
1 <= nums.length <= 3 10^4
-100 <= nums[i] <= 100
0 <= index < nums.length
-100 <= val <= 100
0 <= left <= right < nums.length
调用 update 和 sumRange 方法次数不大于 3 10^4来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/range-sum-query-mutable
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思路
这个题和不同的是,涉及到了数组的更新问题,如果还使用前缀和数组,每更新一处的值就需要影响到前缀和数组的很多值,更新和查询都会达到
#card=math&code=O%28n%29&id=xgv5C)时间复杂度。使用「线段树」或者「树状数组」可以将更新和查询操作都降到
#card=math&code=O%28logn%29&id=gwg4b)复杂度。真正去学就发现也没有那么难,入门可以参考liweiwei的「文章」和「视频」。
目前我对「线段树」或者「树状数组」只是有一个基本的认识,了解了两者的特点和使用场景。比如「线段树」是一棵「平衡二叉树」(不一定是完全二叉树),「树状数组」是一棵「多叉树」,「线段树」可以用在「区间和查询」(还可以处理区间最值)和「单点更新」问题中,「树状数组」可以处理「前缀和查询」和「单点更新」问题,「树状数组」通过前缀和相减也可以获得区间和。总的来说,树状数组有的功能线段树都有,但反过来就不成立,且线段树代码复杂很多。两者都不适用于区间中有「添加和删除」元素的操作。
需要多应用才能更好地掌握这类进阶的数据结构,这里就不多做介绍了。至于本题,套用「线段树」或者「树状数组」的模板即可解答。
代码
线段树
class NumArray {
private SegmentTree<Integer> segmentTree;
public NumArray(int[] nums) {
Integer[] data = new Integer[nums.length];
for (int i = 0; i < nums.length; i++) {
data[i] = nums[i];
}
segmentTree = new SegmentTree<>(data, (a, b) -> a + b);
}
public void update(int i, int val) {
segmentTree.set(i, val);
}
public int sumRange(int i, int j) {
return segmentTree.query(i, j);
}
}
interface Merge<E> {
E merge(E e1, E e2);
}
class SegmentTree<E> {
private E[] tree;
private E[] data;
private Merge<E> merge;
public SegmentTree(E[] arr, Merge<E> merge) {
this.merge = merge;
data = (E[]) new Object[arr.length];
for (int i = 0; i < arr.length; i++) {
data[i] = arr[i];
}
tree = (E[]) new Object[4 * arr.length];
buildSegmentTree(0, 0, arr.length - 1);
}
private void buildSegmentTree(int treeIndex, int l, int r) {
if (l == r) {
tree[treeIndex] = data[l];
return;
}
int mid = l + (r - l) / 2;
int leftChild = leftChild(treeIndex);
int rightChild = rightChild(treeIndex);
buildSegmentTree(leftChild, l, mid);
buildSegmentTree(rightChild, mid + 1, r);
tree[treeIndex] = merge.merge(tree[leftChild], tree[rightChild]);
}
public E query(int dataL, int dataR) {
if (dataL < 0 || dataL >= data.length || dataR < 0 || dataR >= data.length || dataL > dataR) {
throw new IllegalArgumentException("Index is illegal.");
}
return query(0, 0, data.length - 1, dataL, dataR);
}
private E query(int treeIndex, int l, int r, int dataL, int dataR) {
if (l == dataL && r == dataR) {
return tree[treeIndex];
}
int mid = l + (r - l) / 2;
int leftChildIndex = leftChild(treeIndex);
int rightChildIndex = rightChild(treeIndex);
if (dataR <= mid) {
return query(leftChildIndex, l, mid, dataL, dataR);
}
if (dataL >= mid + 1) {
return query(rightChildIndex, mid + 1, r, dataL, dataR);
}
E leftResult = query(leftChildIndex, l, mid, dataL, mid);
E rightResult = query(rightChildIndex, mid + 1, r, mid + 1, dataR);
return merge.merge(leftResult, rightResult);
}
public void set(int dataIndex, E val) {
if (dataIndex < 0 || dataIndex >= data.length) {
throw new IllegalArgumentException("Index is illegal.");
}
data[dataIndex] = val;
set(0, 0, data.length - 1, dataIndex, val);
}
private void set(int treeIndex, int l, int r, int dataIndex, E val) {
if (l == r) {
tree[treeIndex] = val;
return;
}
int leftTreeIndex = leftChild(treeIndex);
int rightTreeIndex = rightChild(treeIndex);
int mid = l + (r - l) / 2;
if (dataIndex >= mid + 1) {
set(rightTreeIndex, mid + 1, r, dataIndex, val);
}
if (dataIndex <= mid) {
set(leftTreeIndex, l, mid, dataIndex, val);
}
tree[treeIndex] = merge.merge(tree[leftTreeIndex], tree[rightTreeIndex]);
}
public int leftChild(int index) {
return 2 * index + 1;
}
public int rightChild(int index) {
return 2 * index + 2;
}
}
/**
* Your NumArray object will be instantiated and called as such:
* NumArray obj = new NumArray(nums);
* obj.update(index,val);
* int param_2 = obj.sumRange(left,right);
*/
树状数组
class NumArray {
FenwickTree fenwickTree;
int[] nums;
public NumArray(int[] nums) {
this.nums = nums;
fenwickTree = new FenwickTree(nums);
}
public void update(int index, int val) {
fenwickTree.update(index + 1, val - nums[index]);
nums[index] = val;
}
public int sumRange(int left, int right) {
return fenwickTree.query(right + 1) - fenwickTree.query(left);
}
}
public class FenwickTree {
private int[] tree;
private int len;
public FenwickTree(int n) {
this.len = n;
tree = new int[n + 1];
}
public FenwickTree(int[] nums) {
this.len = nums.length;
tree = new int[this.len + 1];
for (int i = 1; i <= len; i++) {
update(i, nums[i - 1]);
}
}
public void update(int i, int delta) {
while (i <= len) {
tree[i] += delta;
i += lowbit(i);
}
}
public int query(int i) {
int sum = 0;
while (i > 0) {
sum += tree[i];
i -= lowbit(i);
}
return sum;
}
public static int lowbit(int x) {
return x & (-x);
}
}