334. Increasing Triplet Subsequence 发表于 2022-02-04 Traverse123456789101112131415161718192021222324class Solution { public boolean increasingTriplet(int[] nums) { int[] leftMin = new int[nums.length]; // 左侧至当前下标(含)的最小值 leftMin[0] = nums[0]; for (int i = 1; i < nums.length; i++) { leftMin[i] = Math.min(nums[i], leftMin[i - 1]); } int[] rightMax = new int[nums.length]; // 右侧至当前下标(含)的最大值 rightMax[rightMax.length - 1] = nums[nums.length - 1]; for (int i = nums.length - 2; i >= 0; i--) { rightMax[i] = Math.max(nums[i], rightMax[i + 1]); } for (int i = 1; i < nums.length - 1; i++) { // 注意此处比较的为 leftMin[i - 1] 与 rightMax[i + 1] if (nums[i] > leftMin[i - 1] && nums[i] < rightMax[i + 1]) { return true; } } return false; }} Greedy123456789101112131415161718192021class Solution { public boolean increasingTriplet(int[] nums) { if (nums.length < 3) { return false; } int first = nums[0], second = Integer.MAX_VALUE; for (int i = 1; i < nums.length; i++) { int num = nums[i]; if (num > second) { return true; } else if (num > first) { second = num; } else { first = num; } } return false; }} Reference334. Increasing Triplet Subsequence