leetcode筆記：Range Sum Query - Mutable

leetcode筆記：Range Sum Query - Mutable

一. 題目描述

Given an integer array nums, find the sum of the elements between indices i and j (i ≤ j), inclusive.

The update(i, val) function modifies nums by updating the element at index i to val.

Example:
Given nums = [1, 3, 5]

sumRange(0, 2) -> 9
update(1, 2)
sumRange(0, 2) -> 8

Note:
The array is only modifiable by the update function.
You may assume the number of calls to update and sumRange function is distributed evenly.

二. 題目分析

題目在Range Sum Query - Immutable一題的基礎上增加的難度，要求在輸入數組nums後，能夠修改數組的元素，每次只修改一個元素。同樣要實現求數組的某個區間和的功能。

如果在一題的做法上稍加改進，是可以實現功能的，但是會超時。

這時閱讀了一些文章後才發現原來經典的做法（包括前一題）是使用樹狀數組來維護這個數組nums，其插入和查詢都能做到O(logn)的復雜度，十分巧妙。

三. 示例代碼

// 超時
class NumArray {
public:
    NumArray(vector &nums) {
        if (nums.empty()) return;
        else
        {
            sums.push_back(nums[0]);
            //求得給定數列長度
            int len = nums.size();
            for (int i = 1; i < len; ++i)
                sums.push_back(sums[i - 1] + nums[i]);
        }
    }

    void update(int i, int val) {
        for (int k = i; k < nums.size(); ++k)
            sums[k] += (val - nums[i]);
        nums[i] = val;
    }

    int sumRange(int i, int j) {
        return sums[j] - sums[i - 1];
    }

private:
    vector nums;
    //存儲數列和
    vector sums;
};


// Your NumArray object will be instantiated and called as such:
// NumArray numArray(nums);
// numArray.sumRange(0, 1);
// numArray.update(1, 10);
// numArray.sumRange(1, 2);

// 使用樹狀數組實現的代碼AC，復雜度為O(logn)
class NumArray {
private:
    vector c;
    vector m_nums;
public:
    NumArray(vector &nums) {
        c.resize(nums.size() + 1);
        m_nums = nums;
        for (int i = 0; i < nums.size(); i++){
            add(i + 1, nums[i]);
        }
    }

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

    void add(int pos, int value){
        while (pos < c.size()){
            c[pos] += value;
            pos += lowbit(pos);
        }
    }
    int sum(int pos){
        int res = 0;
        while (pos > 0){
            res += c[pos];
            pos -= lowbit(pos);
        }
        return res;
    }

    void update(int i, int val) {
        int ori = m_nums[i];
        int delta = val - ori;
        m_nums[i] = val;
        add(i + 1, delta);
    }

    int sumRange(int i, int j) {
        return sum(j + 1) - sum(i);
    }
};
// Your NumArray object will be instantiated and called as such:
// NumArray numArray(nums);
// numArray.sumRange(0, 1);
// numArray.update(1, 10);
// numArray.sumRange(1, 2);

四. 小結

之前只是聽過，這是第一次使用樹狀數組，這是維護數組的一個很好的思想，需要深入學習！