【一次过】Lintcode 943. Range Sum Query - Immutable

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

样例

Given nums = [-2, 0, 3, -5, 2, -1]

sumRange(0, 2) -> 1
sumRange(2, 5) -> -1
sumRange(0, 5) -> -3

注意事项

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

---

解题思路1：

常规思路就是直接遍历相加，可是这样时间复杂度高。

class NumArray {

    private int[] res;

    public NumArray(int[] nums) {
        res = Arrays.copyOf(nums , nums.length);
    }
    
    public int sumRange(int i, int j) {
        int sum = 0;
        
        for(int k=i ; k<=j ; k++)
            sum += res[k];
        
        return sum;
    }
}

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

解题思路2：

由于题目中说明sumRange函数会多次调用，所以此函数的时间复杂度不宜过高，需要将此函数的复杂度转移到构造函数中，我们可以假设状态转换res[i]表示到0到i的元素和，这里求nums中i到j的元素和就可以转换为res中res[j+1]与res[i]的差。

class NumArray {

    private int[] res;

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

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