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

一刷
题解：
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。

public class NumArray {
    int[] dp;

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

/**
 * Your NumArray object will be instantiated and called as such:
 * NumArray obj = new NumArray(nums);
 * int param_1 = obj.sumRange(i,j);
 */

二刷
同上

class NumArray {

    int[] sum;
    public NumArray(int[] nums) {
        sum = new int[nums.length];
        int res = 0;
        for(int i=0; i<nums.length; i++){
            res += nums[i];
            sum[i] = res;
        }
    }
    
    public int sumRange(int i, int j) {
        if(i==0) return sum[j];
        else return sum[j] - sum[i-1];
    }
}

/**
 * Your NumArray object will be instantiated and called as such:
 * NumArray obj = new NumArray(nums);
 * int param_1 = obj.sumRange(i,j);
 */