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_307_NumArray_2.java
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_307_NumArray_2.java
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package pp.arithmetic.leetcode;
/**
* Created by wangpeng on 2018/9/29.
* 307.区域和检索 - 数组可修改
* <p>
* 给定一个整数数组 nums,求出数组从索引 i 到 j (i ≤ j) 范围内元素的总和,包含 i, j 两点。
* <p>
* update(i, val) 函数可以通过将下标为 i 的数值更新为 val,从而对数列进行修改。
* <p>
* 示例:
* <p>
* Given nums = [1, 3, 5]
* <p>
* sumRange(0, 2) -> 9
* update(1, 2)
* sumRange(0, 2) -> 8
* 说明:
* <p>
* 数组仅可以在 update 函数下进行修改。
* 你可以假设 update 函数与 sumRange 函数的调用次数是均匀分布的。
*
* @see <a href="https://leetcode-cn.com/problems/range-sum-query-mutable/description/">range-sum-query-mutable</a>
*/
public class _307_NumArray_2 {
public static void main(String[] args) {
NumArray numArray = new NumArray(new int[]{1, 3, 5});
System.out.println(numArray.sumRange(0, 2));
numArray.update(1, 2);
System.out.println(numArray.sumRange(0, 2));
}
/**
* 换一种实现,是复杂度能到O(logn)-->线段树
* update复杂度O(logn)
* sum复杂度O(logn)
*/
private static class NumArray {
int[] nums;
int[] result;
public NumArray(int[] nums) {
this.nums = nums;
result = new int[nums.length * 4];
buildSegmentTree(nums, result, 0, 0, nums.length - 1);
}
private void buildSegmentTree(int[] nums,
int[] result,
int position,
int left,
int right) {
if (left == right) {
result[position] = nums[left];
return;
}
int mid = (left + right) / 2;
buildSegmentTree(nums, result, position * 2 + 1, left, mid);
buildSegmentTree(nums, result, position * 2 + 2, mid + 1, right);
result[position] = result[position * 2 + 1] + result[position * 2 + 2];
}
public void update(int i, int val) {
updateSegmentTree(result, 0, 0, nums.length - 1, i, val);
}
private void updateSegmentTree(int[] result,
int pos,
int left,
int right,
int index,
int newValue) {
if (left == right && left == index) {
result[pos] = newValue;
return;
}
int mid = (left + right) / 2;
if (index <= mid) {
updateSegmentTree(result, pos * 2 + 1, left, mid, index, newValue);
} else {
updateSegmentTree(result, pos * 2 + 2, mid + 1, right, index, newValue);
}
result[pos] = result[pos * 2 + 1] + result[pos * 2 + 2];
}
public int sumRange(int i, int j) {
return sumSegmentTree(result, 0, 0, nums.length - 1, i, j);
}
private int sumSegmentTree(int[] result,
int pos,
int left,
int right,
int pleft,
int pright) {
if (left > pright || right < pleft) {
return 0;
}
if (pleft <= left && pright >= right) {
return result[pos];
}
int mid = (left + right) / 2;
return sumSegmentTree(result, pos * 2 + 1, left, mid, pleft, pright) +
sumSegmentTree(result, pos * 2 + 2, mid + 1, right, pleft, pright);
}
}
}