300. The longest increasing subsequence

• dp time complexity of O (N ^ 2)
  

    public int lengthOfLIS(int[] nums) {
        int[] length = new int[nums.length]; // length[i] 代表以nums[i]结尾的最长上升子序列的长度
        for (int i = 0; i < nums.length; i++) {
            length[i] = 1;
            for (int j = 0; j < i; j++) {
                if(nums[i] > nums[j]) {
                    if (length[j] + 1 > length[i]) {
                        length[i] = length[j] + 1;
                    }
                }
            }
        }

        int result = 0;
        for (int len : length) {
            result = Math.max(result, len);
        }
        return result;
    }

• dp + binary search time complexity of O (N * lg (N))

    public int lengthOfLIS(int[] nums) {
        int[] smallestTails = new int[nums.length];
        int realSize = 0;
        for (int i = 0; i < nums.length; i++) {
            int index = Arrays.binarySearch(smallestTails, 0, realSize, nums[i]);
            if (index < 0) {
                int insertIndex = -index - 1; // insertIndex的逻辑由Arrays.binarySearch方法返回值的形式决定。
                smallestTails[insertIndex] = nums[i];
                realSize = Math.max(realSize, insertIndex + 1); // realSize 代表 smallestTails 不为0的部分的长度。
            } else {
                smallestTails[index] = nums[i];
            }
        }
        return realSize;
    }