【 NO.1 至少在两个数组中出现的值】
解题思路
签到题。
代码展示
class Solution {
public List
int[] count1 = new int[200];
int[] count2 = new int[200];
int[] count3 = new int[200];
for (int n : nums1) {
count1[n] = 1;
}
for (int n : nums2) {
count2[n] = 1;
}
for (int n : nums3) {
count3[n] = 1;
}
List result = new ArrayList<>();
for (int i = 0; i < 200; i++) {
if (count1[i] + count2[i] + count3[i] >= 2) {
result.add(i);
}
}
return result; }
}
图片
【 NO.2 获取单值网格的最小操作数】
解题思路
网格可以转换成一维数组，然后排序。
不返回 -1 的条件就是：任意两个元素的差都是 x 的整倍数，我们只需要检查相邻元素即可。
然后枚举最终的唯一值，可以利用前缀、后缀和快速计算出把所有数字变成该值的操作次数。
代码展示
class Solution {
public int minOperations(int[][] grid, int x) {
int[] arr = new int[grid.length * grid[0].length];
for (int i = 0; i < grid.length; i++) {
for (int j = 0; j < grid[0].length; j++) {
arr[i * grid[0].length + j] = grid[i][j];
}
}
Arrays.sort(arr);
for (int i = 1; i < arr.length; i++) {
if ((arr[i] - arr[i - 1]) % x != 0) {
return -1;
}
}
int suffixSum = Arrays.stream(arr).sum();
int prefixSum = 0;
int result = suffixSum / x;
for (int i = 0; i < arr.length; i++) {
suffixSum -= arr[i];
result = Math.min(result, calc(prefixSum, suffixSum, i, arr.length - i - 1, arr[i], x));
prefixSum += arr[i];
}
return result;}
private int calc(int prefixSum, int suffixSum, int prefixNum, int suffixNum, int target, int step) {
return (prefixNum * target - prefixSum) / step + (suffixSum - suffixNum * target) / step;}
}
图片
【 NO.3 股票价格波动】
解题思路
使用两个 TreeMap 即可，一个储存时间戳到价格的映射，一个储存价格出现了多少次。
代码展示
class StockPrice {
TreeMap
TreeMap
public StockPrice() {
timestampToPrice = new TreeMap<>();
priceCount = new TreeMap<>();}
public void update(int timestamp, int price) {
if (timestampToPrice.containsKey(timestamp)) {
int oldPrice = timestampToPrice.get(timestamp);
int count = priceCount.get(oldPrice);
if (count == 1) {
priceCount.remove(oldPrice);
} else {
priceCount.put(oldPrice, count - 1);
}
}
timestampToPrice.put(timestamp, price);
priceCount.put(price, priceCount.getOrDefault(price, 0) + 1);}
public int current() {
return timestampToPrice.lastEntry().getValue();}
public int maximum() {
return priceCount.lastKey();}
public int minimum() {
return priceCount.firstKey();}
}
图片
【 NO.4 将数组分成两个数组并最小化数组和的差】
解题思路
枚举 + 双指针，具体思路见代码注释。
代码展示
class Solution {
public int minimumDifference(int[] nums) {
int half = nums.length / 2;
int[] half1 = Arrays.copyOf(nums, half);
int[] half2 = new int[nums.length - half];
System.arraycopy(nums, half, half2, 0, half2.length);
// sums1[i] 表示从 half1 中选出 i 个数字得到的和的所有情况，并且从小到大排序
List> sums1 = getSums(half1);
List> sums2 = getSums(half2);
int sum = Arrays.stream(nums).sum();
int result = 0x3f3f3f3f;
// 枚举从 half1 中选出 select 个，则需要从 half2 中选出 half - select 个
for (int select = 0; select <= half; select++) {
List half1Sums = sums1.get(select);
List half2Sums = sums2.get(half - select);
// 从 half1Sums 和 half2Sums 中各选出一个数字，使得它们的和最接近 sum / 2
int i = 0, j = half2Sums.size() - 1;
result = Math.min(result, Math.abs(sum - (half1Sums.get(i) + half2Sums.get(j)) * 2));
for (; i < half1Sums.size(); i++) {
while (j > 0 && Math.abs(sum - (half1Sums.get(i) + half2Sums.get(j - 1)) * 2) <= Math.abs(sum - (half1Sums.get(i) + half2Sums.get(j)) * 2)) {
j--;
}
result = Math.min(result, Math.abs(sum - (half1Sums.get(i) + half2Sums.get(j)) * 2));
}
}
return result; }
// getSums 求出 nums 的所有子集的和
// 返回 List
// 其中 sums[i] 表示 nums 的所有大小为 i 的子集的和
// 去重并排序
private List
int n = nums.length;
List> set = new ArrayList<>();
List> sums = new ArrayList<>();
for (int i = 0; i <= n; i++) {
sums.add(new ArrayList<>());
set.add(new HashSet<>());
}
for (int i = 0; i < (1 << n); i++) {
int sum = 0;
int num = 0;
for (int j = 0; j < n; j++) {
if ((i & (1 << j)) != 0) {
sum += nums[j];
num++;
}
}
if (!set.get(num).contains(sum)) {
set.get(num).add(sum);
sums.get(num).add(sum);
}
}
for (int i = 0; i < n; i++) {
Collections.sort(sums.get(i));
}
return sums; }
}