Leetcode Range Sum Query - Immutable

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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：

最简单的dynamic Programming.  Complexity: NumArray O(n)   sumRange O(1)

为了解决边界问题， 建立accu = new int[nums.length+1]   accu[i]表示nums[0]+...+ nums[i-1]

recurrence formula：accu[i] = accu[i-1] + nums[i-1]     base case: accu[0]= 0

Java code:

public class NumArray {
    private int[] accu;

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

    public int sumRange(int i, int j) {
        return accu[j+1] - accu[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:

1. https://leetcode.com/discuss/68725/5-lines-c-4-lines-python

2. https://leetcode.com/discuss/69081/3ms-java-solution-with-array-no-special-i-0-check

Leetcode Range Sum Query - Immutable

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原文地址：http://www.cnblogs.com/anne-vista/p/4963684.html