303. Range Sum Query - Immutable

题目：

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.

链接： http://leetcode.com/problems/range-sum-query-immutable/

题解：

给定数组，求range之内的数组和。 我们可以新建一个dp数组来保存数据，dp[i]表示 0到i-1这i个数的sum，每次我们就可以在使用O(n)初始化之后，用O(1)的时间得到搜索结果了。

Time Complexity - O(n)， Space Compleixty - O(1)。

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

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


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

Reference:

303. Range Sum Query - Immutable

原文：http://www.cnblogs.com/yrbbest/p/5050025.html