https://leetcode.com/problems/range-sum-query-immutable/
多次计算[i, j]数组和
空间复杂度O(1)即可,dp保存[0, i]的和。此处又用到dp多开一位的trick
public class NumArray {
int[] dp;
public NumArray(int[] nums) {
dp = new int[nums.length + 1];
for (int i = 1; i <= nums.length; i++) {
dp[i] = dp[i - 1] + nums[i - 1];
}
}
public int sumRange(int i, int j) {
return dp[j + 1] - dp[i];
}
}
/**
* Your NumArray object will be instantiated and called as such:
* NumArray obj = new NumArray(nums);
* int param_1 = obj.sumRange(i,j);
*/