题目

303. Range Sum Query - Immutable

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

Given nums = [-2, 0, 3, -5, 2, -1]
sumRange(0, 2) -> 1
sumRange(2, 5) -> -1
sumRange(0, 5) -> -3

Constraints:

You may assume that the array does not change.

There are many calls to sumRange function.

1 穷举解法

发现速度很慢，因为约束说了，这是要重复使用的。只跑赢了5%的提交。

public class NumArray {private int[] nums;public NumArray(int[] nums) {this.nums = nums;}public int sumRange(int i, int j) {int sum = 0;for (; i <= j && j < nums.length; i++) {sum += nums[i];}return sum;}}

2 Cache的方式解法

因为题目强调了，sumRange方法需要被多次调用，那么一定要有缓存，才能达到O(1)的速度。

这里有个公式sums[i + 1] = sums[i] + nums[i]可以把前面多少个数的和缓存起来，求区间数据只要用return sums[j + 1] - sums[i]; 就可以O(1)算出来。

public class NumArray {private int[] sums;public NumArray(int[] nums) {sums = new int[nums.length + 1];for (int i = 0; i < nums.length; i++) {sums[i + 1] = sums[i] + nums[i];}}public int sumRange(int i, int j) {return sums[j + 1] - sums[i];}}

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算法：求数组区间的和Range Sum Query - Immutable

题目

1 穷举解法

2 Cache的方式解法

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