樹狀數組 計算 任意連續N個值的和的時間複雜度爲Log(n) 修改也是Log(n)編程
而普通數組修改是O(1) 計算和是O(n)數組
具體定義能夠看這裏:http://zh.wikipedia.org/zh-cn/%E6%A0%91%E7%8A%B6%E6%95%B0%E7%BB%84this
或者看這個Blog:http://dongxicheng.org/structure/binary_indexed_tree/spa
這東西剛恰好能夠解決 編程之美里面的 1.7光影切割問題code
using System; using System.Collections.Generic; using System.Linq; using System.Text; using System.Threading.Tasks; namespace Clover.Algoritms.DataStructure { public class TreeArray { private double[] items; private double[] data; public TreeArray(double[] data) { if (data == null || data.Length == 0) throw new ArgumentNullException("data"); this.items = new double[data.Length]; this.data = data; for (int i = 1; i <= items.Length; i++) { int start = i - Lowbit(i); double sum = 0; while (start < i) { sum += data[start]; start++; } items[i - 1] = sum; } } public double Sum(int k) { double ret = 0; while (k > 0) { ret += items[k - 1]; k -= Lowbit(k); } return ret; } public void Update(int k, int value) { int x = k - 1; var oldValue = this.data[x]; this.data[x] = value; for (int i = x; i < items.Length; i += Lowbit(i + 1)) { items[i] = items[i] - oldValue + value; } } public static int Lowbit(int i) { return i & -i; } } }