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

 第二遍方法（更好）：定义prefix sum array length: nums.length+1

 1 public class NumArray {
 2     int[] arr;
 3 
 4     public NumArray(int[] nums) {
 5         if (nums.length == 0) return;
 6         arr = new int[nums.length+1];
 7         for (int i=0; i<nums.length; i++) {
 8             arr[i+1] = arr[i] + nums[i];
 9         }
10     }
11 
12     public int sumRange(int i, int j) {
13         return arr[j+1]-arr[i];
14     }
15 }
16 
17 
18 // Your NumArray object will be instantiated and called as such:
19 // NumArray numArray = new NumArray(nums);
20 // numArray.sumRange(0, 1);
21 // numArray.sumRange(1, 2);

第一遍方法：定义prefix sum length 为nums.length, 这样sumRange要讨论i=0的情况

 1 public class NumArray {
 2     int[] leftSums;
 3 
 4     public NumArray(int[] nums) {
 5         leftSums = new int[nums.length];
 6         if (nums.length == 0) return;
 7         leftSums[0] = nums[0];
 8         for (int i=1; i<nums.length; i++) {
 9             leftSums[i] = leftSums[i-1] + nums[i];
10         }
11     }
12 
13     public int sumRange(int i, int j) {
14         return i==0? leftSums[j] : leftSums[j] - leftSums[i-1];
15     }
16 }
17 
18 
19 // Your NumArray object will be instantiated and called as such:
20 // NumArray numArray = new NumArray(nums);
21 // numArray.sumRange(0, 1);
22 // numArray.sumRange(1, 2);

 

转载于:https://www.cnblogs.com/EdwardLiu/p/5084606.html