1.leetcode_1832_Check_if_the_Sentence_Is_Pangram
这一题比较直白，count 一下每个字母出现的次数即可。看独立字母数有没有到 26 即可。
// AC
// Runtime: 5 ms, faster than 25.00% of Java online submissions for Check if the Sentence Is Pangram.
// Memory Usage: 39 MB, less than 25.00% of Java online submissions for Check if the Sentence Is Pangram.
class Solution {
public boolean checkIfPangram(String sentence) {
int size = sentence.length();
Map<String, Integer> record = new HashMap<>();
for(int i = 0; i < size; i++) {
String temp = String.valueOf(sentence.charAt(i));
record.merge(temp, 1, Integer::sum);
}
return record.size() == 26;
}
}
2.leetcode_1833_Maximum_Ice_Cream_Bars
贪心法：先排序，然后从小到大累加。如果累加和恰好等于 k，则返回个数。如果恰号大于 k，上一个累加和恰小于 k，则返回（当前累加计数 - 1）
// AC:
// Runtime: 28 ms, faster than 100.00% of Java online submissions for Maximum Ice Cream Bars.
// Memory Usage: 55.9 MB, less than 25.00% of Java online submissions for Maximum Ice Cream Bars.
class Solution {
public int maxIceCream(int[] costs, int coins) {
Arrays.sort(costs);
int sum = 0;
for(int i = 0; i < costs.length; i++) {
sum += costs[i];
if (sum == coins) {
return i + 1;
}
if (sum > coins) {
if (sum == 0) {
return 0;
} else {
return i;
}
}
}
return costs.length;
}
}
3.leetcode_1834_Single-Threaded_CPU
用优先级队列模拟当前 CPU 可选任务池，每次循环将小于当前 time 的任务插入优先级队列（以执行时间越短越靠前做优先级排序依据），
从队列中 pull 出满足小于当前 time 且执行之间最短的 task
// AC:
// Runtime: 103 ms, faster than 57.14% of Java online submissions for Single-Threaded CPU.
// Memory Usage: 96.7 MB, less than 14.29% of Java online submissions for Single-Threaded CPU.
// 思路：用优先级队列，模拟当前 CPU 可选的任务池，
//
class Solution {
public int[] getOrder(int[][] tasks) {
int taskCount = tasks.length;
int[] ret = new int[taskCount];
int[][] taskRecord = new int[taskCount][3];
for(int i = 0; i < taskCount; i++) {
taskRecord[i][0] = i;
taskRecord[i][1] = tasks[i][0];
taskRecord[i][2] = tasks[i][1];
}
Arrays.sort(taskRecord, (a, b) -> (a[1] - b[1]));
// 按执行时间越短排越前
PriorityQueue<int[]> queue = new PriorityQueue<int[]>((a, b) -> a[2] == b[2] ? a[0] - b[0] : a[2] - b[2]);
int time = 0, step = 0, timeStep = 0;
while (step < taskCount) {
while (timeStep < taskCount && taskRecord[timeStep][1] <= time) {
queue.offer(taskRecord[timeStep++]);
}
if (queue.isEmpty()) {
time = taskRecord[timeStep][1];
continue;
}
int[] curTask = queue.poll();
ret[step++] = curTask[0];
time += curTask[2];
}
return ret;
}
}
4.leetcode_1835_Find_XOR_Sum_of_All_Pairs_Bitwise_AND
(当时没想出来！)
位运算技巧：(a1 ^ a2) & (b1 ^ b2) = (a1 & b1) ^ (a1 & b2) ^ (a2 & b1) ^ (a2 & b2)
// AC:
// Runtime: 1 ms, faster than 100.00% of Java online submissions for Find XOR Sum of All Pairs Bitwise AND.
// Memory Usage: 52.3 MB, less than 25.00% of Java online submissions for Find XOR Sum of All Pairs Bitwise AND.
//
// 思路：这个运算场景满足结合律： (a1 ^ a2) & (b1 ^ b2) = (a1 & b1) ^ (a1 & b2) ^ (a2 & b1) ^ (a2 & b2)
// 复杂度：T：O(m + n), S:O(1)
class Solution {
public int getXORSum(int[] arr1, int[] arr2) {
int ret1 = 0, ret2 = 0;
for(int i: arr1) {
ret1 ^= i;
}
for(int i: arr2) {
ret2 ^= i;
}
return ret1 & ret2;
}
}