No.1
Check if the Sentence Is Pangram
解题思路
简单遍历字符串判断
非英文字母字符直接返回false
否则记录不同字符的个数是否满足26个
代码展示
public boolean checkIfPangram(String sentence) {
if (sentence == null || sentence.length() == 0) {
return false;
}
Set
for (char c : sentence.toCharArray()) {
if (c - 'a' < 0 || c - 'a' >= 26) {
return false;
}
set.add(c);
}
return set.size() == 26;
}
No.2 Maximum Ice Cream Bars
解题思路
贪心的去买雪糕
无需看成背包问题，TLE or MLE
代码展示
public int maxIceCream(int[] costs, int coins) {
if (costs == null || costs.length == 0) {
return 0;
}
int n = costs.length;
int res = 0;
// 排序从小到大买即可
Arrays.sort(costs);
for (int cost : costs) {
if (coins > cost) {
res += 1;
coins -= cost;
}
else {
return res;
}
}
return res;
}
No.3 Single-Threaded CPU
解题思路
堆 + 扫描线结合
添加第三位信息，构建自己的 task model
按时间起始时间从小打到排序
minHeap的操作
按照执行的时长排序
若时长相同按照index先后排序
时间线一直后移，达到开始时间加入堆，同时取出堆顶的task开始执行，记录答案
代码展示
public int[] getOrder(int[][] tasks) {
int n = tasks.length;
// 添加第三维度信息index
int[][] newTasks = new int[n][3];
for (int i = 0; i < tasks.length; i++) {
newTasks[i][0] = tasks[i][0];
newTasks[i][1] = tasks[i][1];
newTasks[i][2] = i;
}
// 按照时间先后排序
Arrays.sort(newTasks, Comparator.comparingInt(x -> x[0]));
PriorityQueue
if (newTasks[x][1] == newTasks[y][1]) {
return newTasks[x][2] - newTasks[y][2];
}
return newTasks[x][1] - newTasks[y][1];
});
int curTime = 1;
int[] res = new int[n];
// task index offer to heap
int i = 0;
// task index add to result
int j = 0;
while (i < n) {
while (i < n && (minHeap.isEmpty() || curTime >= newTasks[i][0])) {
curTime = Math.max(curTime, newTasks[i][0]);
minHeap.offer(i++);
}
// 从heap中拿出执行
if (!minHeap.isEmpty()) {
int[] task = newTasks[minHeap.poll()];
res[j++] = task[2];
curTime += task[1];
}
}
while (!minHeap.isEmpty()) {
res[j++] = newTasks[minHeap.poll()][2];
}
return res;
}
No.4 Find XOR Sum of All Pairs Bitwise AND
解题思路
位运算推导，纯数学题
前置技能：`a ⊕ b = (~a & b) | (a & ~b)`
因此：` (a & b1) ⊕ (a & b2) = [(~a | ~b1) & (a | b2 )] | [[(~a | ~b2) & (a | b1 )]`
因此：`(a & b1) ⊕ (a & b2) = a & (b1 ⊕ b2)`
原题：(a & b1) ⊕ (a1 & b2) (a1 & b3) ⊕(a1 & bm) .......⊕(an & bm)
等同于：[a1 & (b1⊕ b2 ⊕.....bm)] ⊕ [a2 & (b1⊕ b2 ⊕.....bm)] ⊕......⊕ [an & (b1⊕ b2 ⊕.....bm)]
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代码展示
public int getXORSum(int[] arr1, int[] arr2) {
int sumArray2 = 0;
int res = 0;
for (int i : arr2) {
sumArray2 ^= i;
}
for (int i : arr1) {
res ^= (i & sumArray2);
}
return res;
}