LeetCode 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:

You may assume that the array does not change.

There are many calls to sumRange function.

思路：

就是求区间i和j之前的值的和，因为是同一个数组，测试用例为取不同的i和j，如果不考虑效率，每次都是i到j的值相加，但是这样如果对同一个数组测试用例多了之后效率很低；所以要考虑效率问题，求出nums的前i个数的值放到sums中，每次只要调用

sums[j]-sums[i-1]

即可

代码实现：

class NumArray {
public:
NumArray(vector<int> nums) {
if (nums.empty()){//如果nums为空，直接返回
return ;
}
else{
sums.push_back(nums[0]);
//求得给定数列长度
int len = nums.size();
for (int i = 1; i < len; ++i){//分别求nums的前i个和放入sums中
sums.push_back(sums[i - 1] + nums[i]);
}
}
}

int sumRange(int i, int j) {
if (0 == i){
return sums[j];
}
int len = sums.size();

if (i < 0 || i >= len || j < 0 || j >= len || i > j){//如果是这些条件，都是不满足的
return 0;
}
return sums[j] - sums[i-1];//即是sums的前j个和-前i-1个和
}
private:
//sums的第i个值即为nums的前i个数的和（i从0开始）
vector<int> sums;
};

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

输出结果： 179ms