Poison

611. Valid Triangle Number

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class Solution {
    public int triangleNumber(int[] nums) {
        Arrays.sort(nums);

        int count = 0;
        for (int i = 0; i < nums.length - 2; i++) {
            for (int j = i + 1; j < nums.length - 1; j++) {
                int sum = nums[i] + nums[j];
                // 使用二分搜索查找小于 sum 的最大元素，即寻找右边界
                int left = j + 1, right = nums.length;
                while (left < right) {
                    int mid = (left + right) >>> 1;
                    if (nums[mid] < sum) {
                        left = mid + 1; // 尝试向右侧逼近
                    } else if (nums[mid] > sum) {
                        right = mid;
                    } else {
                        right = mid; // 等于 sum 时不合法，向左侧逼近
                    }
                }

                // 跳出时 left = right, 且 left 已经不满足条件，所以只能取 left - 1
                if (left != j + 1) {
                    count += (left - 1) - (j + 1) + 1;
                }
            }
        }

        return count;
    }
}
Two Pointers
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class Solution {
    public int triangleNumber(int[] nums) {
        Arrays.sort(nums);

        int count = 0;
        for (int k = nums.length - 1; k >= 2; k--) {
            int i = 0, j = k - 1;
            while (i < j) {
                if (nums[i] + nums[j] > nums[k]) {
                    // nums[i], nums[i + 1] ... nums[j - 1] 都满足条件
                    count += j - i;
                    j--; // 固定 nums[j] 的已经统计完毕 
                } else {
                    // 此时和小于等于 nums[k], 只能增加最小边的值
                    i++;
                }
            }
        }

        return count;
    }
}
Reference

611. Valid Triangle Number