303. Range Sum Query

#week6

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:

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

分析：

动态规划类型

状态转移方程：f[i] = f[i-1] + nums[i];

代码：

超时版本：

class NumArray {
public:
    int** f;
    NumArray(vector<int> nums) {
        int len = nums.size();
        if (len > 0) {
        f = new int* [len];
        for (int i = 0; i < len; i++) f[i] = new int[len];
        f[0][0] = nums[0];
        for (int i = 0; i < len; i++)
            for (int j = i; j < len; j++) {
                if (i == j) f[i][j] = nums[i];
                else {
                    f[i][j] = f[i][j-1] + nums[j];
                }
            }
        } else {
            f = new int*[1];
            f[0] = new int[1];
            f[0][0] = 0;
        }
    }
    
    int sumRange(int i, int j) {
        return f[i][j];
    }
};

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

修改为一维，不超时：

class NumArray {
public:
    int* f;
    NumArray(vector<int> nums) {
        int len = nums.size();
        if (len > 0) {
            f = new int [len];
            f[0] = nums[0];
            for (int i = 1; i < len; i++)
                f[i] = f[i-1] + nums[i];
        } else {
            f = new int[1];
            f[0] = 0;
        }
    }
    
    int sumRange(int i, int j) {
        return f[j] - f[i-1];
    }
};

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