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阿新 • • 發佈:2022-03-29
https://leetcode-cn.com/problems/range-sum-query-mutable/
給你一個數組 nums ,請你完成兩類查詢。
其中一類查詢要求 更新 陣列nums下標對應的值
另一類查詢要求返回陣列nums中索引left和索引right之間(包含)的nums元素的 和,其中left <= right
實現 NumArray 類:
NumArray(int[] nums) 用整數陣列 nums 初始化物件
void update(int index, int val) 將 nums[index] 的值 更新 為 val
int sumRange(int left, int right) 返回陣列nums中索引left和索引right之間(包含)的nums元素的 和(即,nums[left] + nums[left + 1], ..., nums[right])
示例 1:
輸入:
["NumArray", "sumRange", "update", "sumRange"]
[[[1, 3, 5]], [0, 2], [1, 2], [0, 2]]
輸出:
[null, 9, null, 8]
解釋:
NumArray numArray = new NumArray([1, 3, 5]);
numArray.sumRange(0, 2); // 返回 1 + 3 + 5 = 9
numArray.update(1, 2); // nums = [1,2,5]
numArray.sumRange(0, 2); // 返回 1 + 2 + 5 = 8
方法一:線性樹
class NumArray { int[] nums; int[] tree; int n; public NumArray(int[] nums) { this.nums = nums; n = nums.length; if (n == 0) { return; } tree = new int[n * 4]; buildTree(0, 0, n - 1); } void buildTree(int node, int start, int end) { if (start == end) { tree[node] = nums[start]; return; } int mid = (end + start) / 2 + start; int left = node * 2 + 1; int right = node * 2 + 2; buildTree(left, start, mid); buildTree(right, mid + 1, end); tree[node] = tree[left] + tree[right]; } public void update(int index, int val) { updateTree(index, 0, val, 0, n - 1); } public void updateTree(int idx, int node, int val, int start, int end) { if (start > end) { return; } if (start == end) { nums[idx] = val; tree[node] = val; } else { int mid = (start + end) >> 1; int left = node * 2 + 1; int right = node * 2 + 2; if (idx >= start && idx <= mid) { updateTree(idx, left, val, start, mid); } else { updateTree(idx, right, val, mid + 1, end); } tree[node] = tree[left] + tree[right]; } } public int sumRange(int left, int right) { return query(left, right, 0, 0, n - 1); } public int query(int l, int r, int node, int start, int end) { if (l > end || r < start) { return 0; } if (start == end) { return tree[node]; } if (l <= start && r <= end) { return tree[node]; } else { int mid = (start + end) >> 1; int left = node * 2 + 1; int right = node * 2 + 2; int ls = query(l, r, left, start, mid); int rs = query(l, r, right, end, mid + 1); return ls + rs; } } }