303. 区域和检索 - 数组不可变
给定一个整数数组 nums,求出数组从索引 *i *到 j (i ≤ j) 范围内元素的总和,包含 *i, j *两点。
示例:
给定 nums = [-2, 0, 3, -5, 2, -1],求和函数为 sumRange()
sumRange(0, 2) -> 1
sumRange(2, 5) -> -1
sumRange(0, 5) -> -3
class NumArray {
public:
vector<int> dp;
NumArray(vector<int>& nums) {
if(nums.size()>0) //输入为空
{
int n=nums.size();
dp.resize(n,0);
dp[0]=nums[0];
for(int i=1;i<n;++i)
{
dp[i]=dp[i-1]+nums[i];
}
}
}
int sumRange(int i, int j) {
return i==0?dp[j]:dp[j]-dp[i-1];
}
};
/**
* Your NumArray object will be instantiated and called as such:
* NumArray* obj = new NumArray(nums);
* int param_1 = obj->sumRange(i,j);
*/