区域和检索数组可修改

2021-01-13

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下标对应的值，另一类查询要求返回数组中某个范围内元素的总和。

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) 返回子数组 nums[left, right] 的总和（即，nums[left] + nums[left + 1], ..., nums[right]）

示例：

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 9 ，sum([1,3,5]) = 9
numArray.update(1, 2);   // nums = [1,2,5]
numArray.sumRange(0, 2); // 返回 8 ，sum([1,2,5]) = 9

提示：
1 <= nums.length <= 3 * 104
-100 <= nums[i] <= 100
0 <= index < nums.length
-100 <= val <= 100
0 <= left <= right < nums.length
最多调用 3 * 104 次 update 和 sumRange 方法

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/range-sum-query-mutable
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

树状数组

class NumArray {
    FenwickTree fenwickTree;

    int[] nums = null;

    public NumArray(int[] nums) {
        fenwickTree = new FenwickTree(nums.length);
        int count = 1;
        this.nums = nums;
        for (int num : nums) {
            fenwickTree.update(count, num);
            count++;
        }
    }

    public void update(int index, int val) {

        fenwickTree.update(index + 1, -nums[index]);
        nums[index] = val;
        fenwickTree.update(index + 1, val);
    }

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

    private class FenwickTree {

        int len = 0;
        int[] tree;

        public FenwickTree(int n) {
            this.len = n;
            tree = new int[n + 1];
        }


        public void update(int i, int num) {
            while (i <= len) {
                tree[i] += num;
                i += lowbit(i);
            }
        }

        public int query(int i) {
            int sum = 0;
            while (i > 0) {
                sum += tree[i];
                i -= lowbit(i);
            }
            return sum;
        }


        public int lowbit(int x) {

            return x & (-x);
        }

    }
}

/**
 * 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);
 */

线段树（待填坑）

// 构建线段树
int[] tree;
int n;
public NumArray(int[] nums) {
    if (nums.length > 0) {
        n = nums.length;
        tree = new int[n * 2];
        buildTree(nums);
    }
}
private void buildTree(int[] nums) {
    for (int i = n, j = 0;  i < 2 * n; i++,  j++)
        tree[i] = nums[j];
    for (int i = n - 1; i > 0; --i)
        tree[i] = tree[i * 2] + tree[i * 2 + 1];
}
// 更新线段树
void update(int pos, int val) {
    pos += n;
    tree[pos] = val;
    while (pos > 0) {
        int left = pos;
        int right = pos;
        if (pos % 2 == 0) {
            right = pos + 1;
        } else {
            left = pos - 1;
        }
        // parent is updated after child is updated
        tree[pos / 2] = tree[left] + tree[right];
        pos /= 2;
    }
}

// 线段树求和
public int sumRange(int l, int r) {
    // get leaf with value 'l'
    l += n;
    // get leaf with value 'r'
    r += n;
    int sum = 0;
    while (l <= r) {
        if ((l % 2) == 1) {
           sum += tree[l];
           l++;
        }
        if ((r % 2) == 0) {
           sum += tree[r];
           r--;
        }
        l /= 2;
        r /= 2;
    }
    return sum;
}


作者：LeetCode
链接：https://leetcode-cn.com/problems/range-sum-query-mutable/solution/qu-yu-he-jian-suo-shu-zu-ke-xiu-gai-by-leetcode/
来源：力扣（LeetCode）
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。