思路: 模拟

class Solution {public String triangleType(int[] nums) {// 先判合法性Arrays.sort(nums);if (nums[0] + nums[1] <= nums[2]) return "none";if (nums[0] == nums[1] && nums[1] == nums[2]) {return "equilateral";} else if (nums[0] == nums[1] || nums[1] == nums[2]) {return "isosceles";} else {return "scalene";}}
}

和T4一起讲

思路: 基于map的前缀和应用

这边需要以值作为key, value为最小的前缀和(需向前偏移一位)

更新的时候，需要分类讨论，v为当前值

• $v-k$
• $v+k$

class Solution {public long maximumSubarraySum(int[] nums, int k) {long inf = Long.MIN_VALUE / 10;long res = inf;// 维护最小的前缀和Map<Long, Long> minMap = new HashMap<>();long acc = 0;for (int i = 0; i < nums.length; i++) {long v = nums[i];acc += v;if (minMap.containsKey(v - k)) {res = Math.max(acc - minMap.get(v - k), res);}if (minMap.containsKey(v + k)) {res = Math.max(acc - minMap.get(v + k), res);}// 更新if (!minMap.containsKey(v) || acc - v < minMap.get(v)) {minMap.put(v, acc - v);}}return res == inf ? 0 : res;}}

思路: 二维偏序 + 枚举

对于偏序题，一般先固定一个维度

1. 先按x坐标从小到大排序，
2. 再按照y坐标从大到小排序

因为题目指定左上角，右下角

然后枚举左右端点，check是否满足需求即可。

在枚举的过程中，可以引入

$$单调队列优化$$

实际上只要维护最接近左端点y坐标(严格小于等于)的单变量即可, 递增状态

这样整个时间复杂度可以降为

• 排序$O(nlogn)$
• 枚举左右端点$O(n^2)$

最终为$O(n^2)$

class Solution {public int numberOfPairs(int[][] points) {// 按x从小到大，按y从大到小Arrays.sort(points, Comparator.comparingInt((int[] p) -> p[0]).thenComparingInt(p -> -p[1]));int res = 0;int n = points.length;for (int i = 0; i < n; i++) {// 维护最接近左端点y值的值(严格小于等于)int nearest = Integer.MIN_VALUE;for (int j = i + 1; j < n; j++) {if (points[j][1] <= points[i][1]) {if (points[j][1] > nearest) {res++;nearest = points[j][1];}}}}return res;}
}

• 离散化+二维前缀和 (补充)

这个解法应该更加的直观

class Solution {// 离散化Map<Integer, Integer> discrete(List<Integer> ps) {TreeSet<Integer> range = new TreeSet<>(ps);Map<Integer, Integer> ids = new HashMap<>();int ptr = 0;for (var k: range) {ids.put(k, ptr++);}return ids;}public int numberOfPairs(int[][] points) {int n = points.length;int res = 0;Map<Integer, Integer> xs = discrete(Arrays.stream(points).map(p -> p[0]).collect(Collectors.toList()));Map<Integer, Integer> ys = discrete(Arrays.stream(points).map(p -> p[1]).collect(Collectors.toList()));int h = ys.size(), w = xs.size();int[][] area = new int[h][w];for (int[] p: points) {area[ys.get(p[1])][xs.get(p[0])] = 1;}int[][] pre = new int[h + 1][w + 1];for (int i = 0; i < h; i++) {for (int j = 0; j < w; j++) {pre[i + 1][j + 1] = pre[i + 1][j] + pre[i][j + 1] - pre[i][j] + area[i][j];}}for (int i = 0; i < n; i++) {for (int j = 0; j < n; j++) {if (i == j) continue;if (points[i][0] <= points[j][0] && points[i][1] >= points[j][1]) {int ty = ys.get(points[i][1]), by = ys.get(points[j][1]);int tx = xs.get(points[j][0]), bx = xs.get(points[i][0]);int s = pre[ty + 1][tx + 1] - pre[ty + 1][bx] - pre[by][tx + 1] + pre[by][bx];if (s == 2) {res ++;}}}}return res;}
}