303. Range Sum Query - Immutable
2016-11-21 18:58
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动态规划问题,刚开始想直接写个二维数组,然后存下所有sum[i,j],后来发现测试案例有1W,然后RUN TIME ERROR了
后来发现其实只需要一维数组就可以解决,sum[i]存储i之前所有值的和,求解sum[i,j]=sum[j]-sum[i-1]
还有就是数组为空时返回0,感觉这个测试案例并没有什么卵用
class NumArray {
public:
int*sum;
NumArray(vector<int> &nums) {
if (nums.size() == 0)
sum = NULL;
else
{
sum = new int[nums.size()];
sum[0] = nums[0];
for (int i = 1; i < nums.size(); ++i)
sum[i] = sum[i - 1] + nums[i];
}
}
int sumRange(int i, int j) {
if (!sum)
return 0;
else if (i == 0)
return sum[j];
else
return sum[j] - sum[i - 1];
}
};
后来发现其实只需要一维数组就可以解决,sum[i]存储i之前所有值的和,求解sum[i,j]=sum[j]-sum[i-1]
还有就是数组为空时返回0,感觉这个测试案例并没有什么卵用
class NumArray {
public:
int*sum;
NumArray(vector<int> &nums) {
if (nums.size() == 0)
sum = NULL;
else
{
sum = new int[nums.size()];
sum[0] = nums[0];
for (int i = 1; i < nums.size(); ++i)
sum[i] = sum[i - 1] + nums[i];
}
}
int sumRange(int i, int j) {
if (!sum)
return 0;
else if (i == 0)
return sum[j];
else
return sum[j] - sum[i - 1];
}
};
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