Java [Leetcode 303]Range Sum Query - Immutable

题目描述:

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.

解题思路:

如代码所示。

代码如下:

public class NumArray {
	int[] nums;

    public NumArray(int[] nums) {
        for(int i = 1; i < nums.length; i++){
        	nums[i] += nums[i - 1];
        }
        this.nums = nums;
    }

    public int sumRange(int i, int j) {
        if(i == 0)
        	return nums[j];
        return nums[j] - nums[i - 1];
    }
}

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

  

时间: 2024-10-14 12:10:49

Java [Leetcode 303]Range Sum Query - Immutable的相关文章

leetCode 303. Range Sum Query - Immutable | Dynamic Programming

303. Range Sum Query - Immutable 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:

LeetCode 303. Range Sum Query - Immutable

求数组nums[i,j]的和 思路:另开一sum数组,sum[i]为nums[0,i]的和,所以nums[i,j] = sum[j] - sum[i-1] 1 class NumArray { 2 public: 3 vector<int> sum; 4 NumArray(vector<int> &nums) { 5 sum.resize(nums.size(), 0); 6 sum[0] = nums[0]; 7 int len = nums.size(); 8 for(

leetcode 303. Range Sum Query - Immutable(前缀和)

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: You may assume that the array do

LeetCode:Range Sum Query - Immutable - 数组指定区间内的元素和

1.题目名称 Range Sum Query(数组指定区间内的元素和) 2.题目地址 https://leetcode.com/problems/range-sum-query-immutable/ 3.题目内容 英文:Given an integer array nums, find the sum of the elements between indices i and j (i ≤ j), inclusive. 中文:给定一个数组nums,求出索引i和j之间元素的和,i一定是小于或等于j

303. Range Sum Query - Immutable

303. Range Sum Query - Immutable Total Accepted: 10632 Total Submissions: 44726 Difficulty: Easy 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] sumRang

[leetcode] 303. Range Sum Query &amp;&amp; 304. Range Sum Query 2D - Immutable

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: You may assume that the array do

303. Range Sum Query - Immutable(动态规划)

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

303. Range Sum Query - Immutable 数组范围求和 - 不变

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: You may assume that the array do

303. Range Sum Query - Immutable(动态规划)

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 法一:暴力 1 class NumArray { 2 3 public: 4