No.1最大升序子数组和
解题思路
注意该题目的子数组是连续的。因此枚举起始位置即可。
代码展示
class Solution {
public int maxAscendingSum(int[] nums) {
int res = nums[0];
for (int start = 0; start < nums.length; start++) {
int sum = nums[start];
for (int i = start + 1; i < nums.length; i++) {
if (nums[i] > nums[i - 1]) {
sum += nums[i];
} else {
break;
}
}
res = Math.max(res, sum);
}
return res;
}
}
No.2 积压订单中的订单总数
解题思路
使用 PriorityQueue 或者 TreeMap 维护两类挤压订单即可。
代码展示
class Solution {
public int getNumberOfBacklogOrders(int[][] orders) {
TreeMap
TreeMap
for (var order : orders) {
if (order[2] == 0) {
// buy
while (order[1] > 0) {
var entry = toSell.firstEntry();
if (entry == null || entry.getKey() > order[0]) {
break;
}
int cnt = Math.min(entry.getValue(), order[1]);
order[1] -= cnt;
if (cnt == entry.getValue()) {
toSell.remove(entry.getKey());
} else {
toSell.put(entry.getKey(), entry.getValue() - cnt);
}
// break;
}
if (order[1] > 0) {
toBuy.put(order[0], order[1] + toBuy.getOrDefault(order[0], 0));
}
} else {
// sell
while (order[1] > 0) {
var entry = toBuy.lastEntry();
if (entry == null || entry.getKey() < order[0]) {
break;
}
int cnt = Math.min(entry.getValue(), order[1]);
order[1] -= cnt;
if (cnt == entry.getValue()) {
toBuy.remove(entry.getKey());
} else {
toBuy.put(entry.getKey(), entry.getValue() - cnt);
}
// break;
}
if (order[1] > 0) {
toSell.put(order[0], order[1] + toSell.getOrDefault(order[0], 0));
}
}
}
int res = 0;
for (var entry : toSell.entrySet()) {
res = (res + entry.getValue()) % 1000000007;
}
for (var entry : toBuy.entrySet()) {
res = (res + entry.getValue()) % 1000000007;
}
return res;
}
}
No.3 有界数组中指定下标处的最大值
解题思路
二分答案。当指定下标处的值确定后，贪心地令其他元素尽可能小即可，所以就是等差数列求和。
代码展示
class Solution {
public int maxValue(int n, int index, int maxSum) {
long l = 0, r = maxSum;
while (l + 1 < r) {
long mid = (l + r) / 2;
if (check(mid, index, maxSum, n)) {
l = mid;
} else {
r = mid;
}
}
return (int) (check(r, index, maxSum, n) ? r : l);
}
long sum(long start, long count) {
if (count <= 0) {
return 0;
}
if (start < count) {
return (start + 1) * start / 2 + (count - start);
}
return (start + start - count + 1) * count / 2;
}
boolean check(long value, long index, long maxSum, long n) {
long left = index;
long right = n - index - 1;
return sum(value - 1, left) + sum(value - 1, right) + value <= maxSum;
}
}
No.4 统计异或值在范围内的数对有多少
解题思路
求 [low, high] 区间内的等同于求 [low, inf) 再减去 (high, inf)。
因此问题转换成求异或值大于给定数值的数对数量。使用字典树即可。
代码展示
class Solution {
public int countPairs(int[] nums, int low, int high) {
return (int) (solve(nums, low - 1) - solve(nums, high));
}
static class TrieTree {
TrieTree[] next = new TrieTree[2];
int count = 1;
}
long solve(int[] nums, int m) {
TrieTree trieTree = buildTrieTree(nums);
long result = 0;
for (int i = 0; i < nums.length; i++) {
result += queryTrieTree(trieTree, nums[i], m, 31);
}
return result / 2;
}
long queryTrieTree(TrieTree trieTree, int a, int m, int index) {
if (trieTree == null)
return 0;
TrieTree current = trieTree;
for (int i = index; i >= 0; i--) {
int aDigit = (a >> i) & 1;
int mDigit = (m >> i) & 1;
if (aDigit == 1 && mDigit == 1) {
if (current.next[0] == null)
return 0;
current = current.next[0];
} else if (aDigit == 0 && mDigit == 1) {
if (current.next[1] == null)
return 0;
current = current.next[1];
} else if (aDigit == 1 && mDigit == 0) {
long p = queryTrieTree(current.next[1], a, m, i - 1);
long q = current.next[0] == null ? 0 : current.next[0].count;
return p + q;
} else if (aDigit == 0 && mDigit == 0) {
long p = queryTrieTree(current.next[0], a, m, i - 1);
long q = current.next[1] == null ? 0 : current.next[1].count;
return p + q;
}
}
return 0;
}
TrieTree buildTrieTree(int[] a) {
TrieTree trieTree = new TrieTree();
for (int i = 0; i < a.length; i++) {
TrieTree current = trieTree;
for (int j = 31; j >= 0; j--) {
int digit = (a[i] >> j) & 1;
if (current.next[digit] == null) {
current.next[digit] = new TrieTree();
} else {
current.next[digit].count++;
}
current = current.next[digit];
}
}
return trieTree;
}
}