LC-307.区域和检索-数组可修改

题目描述

leetcode 中等题

给你一个数组 nums ，请你完成两类查询。

1. 其中一类查询要求 更新 数组 nums 下标对应的值
2. 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：

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

树状数组

这是一道树状数组和线段树的模板题。

class NumArray {
    int[] org;
    FenwickTree fenwickTree;

    public NumArray(int[] nums) {
        this.org = nums;
        fenwickTree = new FenwickTree(org);
    }

    public void update(int index, int val) {
        int delta = val - org[index];
        org[index] = val;
        fenwickTree.update(index + 1, delta);
    }

    public int sumRange(int left, int right) {
        return fenwickTree.query(right + 1) - fenwickTree.query(left);
    }
}

class FenwickTree{
    int[] arr;
    int size;

    public FenwickTree(int size) {
        arr = new int[size];
        this.size = size;
    }

    public FenwickTree(int[] org) {
        arr = new int[org.length + 1];
        this.size = arr.length;
        for (int i = 1; i < size; i++) {
            update(i, org[i - 1]);
        }
    }

    public void update(int index, int delta) {
        while (index < size) {
            arr[index] += delta;
            index += lowBit(index);
        }
    }

    public int query(int index) {
        int sum = 0;
        while (index > 0) {
            sum += arr[index];
            index -= lowBit(index);
        }
        return sum;
    }

    private int lowBit(int i) {
        return i & -i;
    }
}

线段树

class NumArray {
    int[] org;
    SegTree segTree;
    
    public NumArray(int[] nums) {
        this.org = nums;
        segTree = new SegTree(nums);
    }

    public void update(int index, int val) {
        int delta = val - org[index];
        org[index] = val;
        segTree.update(index, index, delta, 0, org.length - 1, 1);
    }

    public int sumRange(int left, int right) {
        return segTree.query(left, right, 0, org.length - 1, 1);
    }
}

class SegTree {
    int[] arr;
    int[] lazyTag;
    int[] org;

    public SegTree(int[] org) {
        this.org = org;
        this.arr = new int[org.length * 4];
        this.lazyTag = new int[org.length * 4];
        build(0, org.length - 1, 1);
    }

    private void build(int s, int t, int p) {
        if (s == t) {
            arr[p] = org[s];
            return;
        }
        int mid = s + (t - s >> 1);
        build(s, mid, p * 2);
        build(mid + 1, t, p * 2 + 1);
        arr[p] = arr[p * 2] + arr[p * 2 + 1];
    }

    public void update(int l, int r, int c, int s, int t, int p){
        if (l <= s && r >= t) {
            arr[p] += (t - s + 1) * c;
            lazyTag[p] += c;
            return;
        }
        int mid = s + (t - s >> 1);
        if (lazyTag[p] != 0) {
            arr[p * 2] += (mid - s + 1) * lazyTag[p];
            arr[p * 2 + 1] += (t - mid) * lazyTag[p];
            lazyTag[p * 2] += lazyTag[p];
            lazyTag[p * 2 + 1] += lazyTag[p];
            lazyTag[p] = 0;
        }
        if (l <= mid) {
            update(l, r, c, s, mid, p * 2);
        }
        if (r >= mid + 1) {
            update(l, r, c, mid + 1, t, p * 2 + 1);
        }
        arr[p] = arr[p * 2] + arr[p * 2 + 1];
    }

    public int query(int l, int r, int s, int t, int p){
        if (l <= s && r >= t) {
            return arr[p];
        }
        int mid = s + (t - s >> 1);
        if (lazyTag[p] != 0) {
            arr[p * 2] += (mid - s + 1) * lazyTag[p];
            arr[p * 2 + 1] += (t - mid) * lazyTag[p];
            lazyTag[p * 2] += lazyTag[p];
            lazyTag[p * 2 + 1] += lazyTag[p];
            lazyTag[p] = 0;
        }
        int sum = 0;
        if (l <= mid) {
            sum += query(l, r, s, mid, p * 2);
        }
        if (r >= mid + 1) {
            sum += query(l, r, mid + 1, t, p * 2 + 1);
        }
        return sum;
    }

}