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303. Range Sum Query - Immutable

时间:2019-02-15 22:29:09      阅读:172      评论:0      收藏:0      [点我收藏+]

标签:UNC   between   []   int   ice   numa   cto   any   anti   

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

Example:

Given nums = [-2, 0, 3, -5, 2, -1]

sumRange(0, 2) -> 1
sumRange(2, 5) -> -1
sumRange(0, 5) -> -3

 

Note:

  1. You may assume that the array does not change.
  2. There are many calls to sumRange function.

 

Approach #1:  My code. [C++]

class NumArray {
public:
    NumArray(vector<int> nums) {
        nums_ = nums;
    }
    
    int sumRange(int i, int j) {
        int ans = 0;
        for (int k = i; k <= j; ++k)
            ans += nums_[k];
        return ans;
    }
    
private:
    vector<int> nums_;
};

/**
 * Your NumArray object will be instantiated and called as such:
 * NumArray obj = new NumArray(nums);
 * int param_1 = obj.sumRange(i,j);
 */

  

Approach #2 (Caching) [Java]

 

private int[] sum;

public NumArray(int[] nums) {
    sum = new int[nums.length + 1];
    for (int i = 0; i < nums.length; i++) {
        sum[i + 1] = sum[i] + nums[i];
    }
}

public int sumRange(int i, int j) {
    return sum[j + 1] - sum[i];
}

  

 

303. Range Sum Query - Immutable

标签:UNC   between   []   int   ice   numa   cto   any   anti   

原文地址:https://www.cnblogs.com/ruruozhenhao/p/10386163.html

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