返回介绍

solution / 0300-0399 / 0370.Range Addition / README_EN

发布于 2024-06-17 01:04:01 字数 7372 浏览 0 评论 0 收藏 0

370. Range Addition

中文文档

Description

You are given an integer length and an array updates where updates[i] = [startIdxi, endIdxi, inci].

You have an array arr of length length with all zeros, and you have some operation to apply on arr. In the ith operation, you should increment all the elements arr[startIdxi], arr[startIdxi + 1], ..., arr[endIdxi] by inci.

Return arr _after applying all the_ updates.

 

Example 1:

Input: length = 5, updates = [[1,3,2],[2,4,3],[0,2,-2]]
Output: [-2,0,3,5,3]

Example 2:

Input: length = 10, updates = [[2,4,6],[5,6,8],[1,9,-4]]
Output: [0,-4,2,2,2,4,4,-4,-4,-4]

 

Constraints:

  • 1 <= length <= 105
  • 0 <= updates.length <= 104
  • 0 <= startIdxi <= endIdxi < length
  • -1000 <= inci <= 1000

Solutions

Solution 1

class Solution:
  def getModifiedArray(self, length: int, updates: List[List[int]]) -> List[int]:
    d = [0] * length
    for l, r, c in updates:
      d[l] += c
      if r + 1 < length:
        d[r + 1] -= c
    return list(accumulate(d))
class Solution {
  public int[] getModifiedArray(int length, int[][] updates) {
    int[] d = new int[length];
    for (var e : updates) {
      int l = e[0], r = e[1], c = e[2];
      d[l] += c;
      if (r + 1 < length) {
        d[r + 1] -= c;
      }
    }
    for (int i = 1; i < length; ++i) {
      d[i] += d[i - 1];
    }
    return d;
  }
}
class Solution {
public:
  vector<int> getModifiedArray(int length, vector<vector<int>>& updates) {
    vector<int> d(length);
    for (auto& e : updates) {
      int l = e[0], r = e[1], c = e[2];
      d[l] += c;
      if (r + 1 < length) d[r + 1] -= c;
    }
    for (int i = 1; i < length; ++i) d[i] += d[i - 1];
    return d;
  }
};
func getModifiedArray(length int, updates [][]int) []int {
  d := make([]int, length)
  for _, e := range updates {
    l, r, c := e[0], e[1], e[2]
    d[l] += c
    if r+1 < length {
      d[r+1] -= c
    }
  }
  for i := 1; i < length; i++ {
    d[i] += d[i-1]
  }
  return d
}
/**
 * @param {number} length
 * @param {number[][]} updates
 * @return {number[]}
 */
var getModifiedArray = function (length, updates) {
  const d = new Array(length).fill(0);
  for (const [l, r, c] of updates) {
    d[l] += c;
    if (r + 1 < length) {
      d[r + 1] -= c;
    }
  }
  for (let i = 1; i < length; ++i) {
    d[i] += d[i - 1];
  }
  return d;
};

Solution 2

class BinaryIndexedTree:
  def __init__(self, n):
    self.n = n
    self.c = [0] * (n + 1)

  @staticmethod
  def lowbit(x):
    return x & -x

  def update(self, x, delta):
    while x <= self.n:
      self.c[x] += delta
      x += BinaryIndexedTree.lowbit(x)

  def query(self, x):
    s = 0
    while x:
      s += self.c[x]
      x -= BinaryIndexedTree.lowbit(x)
    return s


class Solution:
  def getModifiedArray(self, length: int, updates: List[List[int]]) -> List[int]:
    tree = BinaryIndexedTree(length)
    for start, end, inc in updates:
      tree.update(start + 1, inc)
      tree.update(end + 2, -inc)
    return [tree.query(i + 1) for i in range(length)]
class Solution {
  public int[] getModifiedArray(int length, int[][] updates) {
    BinaryIndexedTree tree = new BinaryIndexedTree(length);
    for (int[] e : updates) {
      int start = e[0], end = e[1], inc = e[2];
      tree.update(start + 1, inc);
      tree.update(end + 2, -inc);
    }
    int[] ans = new int[length];
    for (int i = 0; i < length; ++i) {
      ans[i] = tree.query(i + 1);
    }
    return ans;
  }
}

class BinaryIndexedTree {
  private int n;
  private int[] c;

  public BinaryIndexedTree(int n) {
    this.n = n;
    c = new int[n + 1];
  }

  public void update(int x, int delta) {
    while (x <= n) {
      c[x] += delta;
      x += lowbit(x);
    }
  }

  public int query(int x) {
    int s = 0;
    while (x > 0) {
      s += c[x];
      x -= lowbit(x);
    }
    return s;
  }

  public static int lowbit(int x) {
    return x & -x;
  }
}
class BinaryIndexedTree {
public:
  int n;
  vector<int> c;

  BinaryIndexedTree(int _n)
    : n(_n)
    , c(_n + 1) {}

  void update(int x, int delta) {
    while (x <= n) {
      c[x] += delta;
      x += lowbit(x);
    }
  }

  int query(int x) {
    int s = 0;
    while (x > 0) {
      s += c[x];
      x -= lowbit(x);
    }
    return s;
  }

  int lowbit(int x) {
    return x & -x;
  }
};

class Solution {
public:
  vector<int> getModifiedArray(int length, vector<vector<int>>& updates) {
    BinaryIndexedTree* tree = new BinaryIndexedTree(length);
    for (auto& e : updates) {
      int start = e[0], end = e[1], inc = e[2];
      tree->update(start + 1, inc);
      tree->update(end + 2, -inc);
    }
    vector<int> ans;
    for (int i = 0; i < length; ++i) ans.push_back(tree->query(i + 1));
    return ans;
  }
};
type BinaryIndexedTree struct {
  n int
  c []int
}

func newBinaryIndexedTree(n int) *BinaryIndexedTree {
  c := make([]int, n+1)
  return &BinaryIndexedTree{n, c}
}

func (this *BinaryIndexedTree) lowbit(x int) int {
  return x & -x
}

func (this *BinaryIndexedTree) update(x, delta int) {
  for x <= this.n {
    this.c[x] += delta
    x += this.lowbit(x)
  }
}

func (this *BinaryIndexedTree) query(x int) int {
  s := 0
  for x > 0 {
    s += this.c[x]
    x -= this.lowbit(x)
  }
  return s
}

func getModifiedArray(length int, updates [][]int) []int {
  tree := newBinaryIndexedTree(length)
  for _, e := range updates {
    start, end, inc := e[0], e[1], e[2]
    tree.update(start+1, inc)
    tree.update(end+2, -inc)
  }
  ans := make([]int, length)
  for i := range ans {
    ans[i] = tree.query(i + 1)
  }
  return ans
}

如果你对这篇内容有疑问,欢迎到本站社区发帖提问 参与讨论,获取更多帮助,或者扫码二维码加入 Web 技术交流群。

扫码二维码加入Web技术交流群

发布评论

需要 登录 才能够评论, 你可以免费 注册 一个本站的账号。
列表为空,暂无数据
    我们使用 Cookies 和其他技术来定制您的体验包括您的登录状态等。通过阅读我们的 隐私政策 了解更多相关信息。 单击 接受 或继续使用网站,即表示您同意使用 Cookies 和您的相关数据。
    原文