[LeetCode] 303. Range Sum Query - Immutable

轉自LeetCode

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.

<Solution>

這題最需要注意的地方是，題目說會呼叫非常多次的 sumRange

所以如果每次再去計算和，很容易TLE

想法如下

• 在內部儲存一個 sum array，定義是 sum[n] = nums[n-1] + sum[n-1]，sum[0] = 0

• 這樣當要計算 nums[i] + ... + nums[j]，就等於 sum[j+1] - sum[i]

code如下

	class NumArray {
	private int[] sumArray;
	public NumArray(int[] nums) {
	sumArray = new int[nums.length+1];
	for(int i = 1; i < sumArray.length; ++i) {
	sumArray[i] = nums[i-1] + sumArray[i-1];
	}
	}
	public int sumRange(int i, int j) {
	return sumArray[j+1] - sumArray[i];
	}
	}
	/**
	* Your NumArray object will be instantiated and called as such:
	* NumArray obj = new NumArray(nums);
	* int param_1 = obj.sumRange(i,j);
	*/

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