[leetcode-300]Longest Increasing Subsequence(java)

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问题描述：这里写链接内容

分析：这道题思路还是比较简单，它要求的是满足这样条件的序列的最大长度，nums[i]

//O(n2)复杂度public class Solution {    public int lengthOfLIS(int[] nums) {        int max = 0;        if(nums.length <= 1)            return nums.length;        for(int i = 0;i<nums.length;i++){            int cur = nums[i];            int count = 0;            int left = i-1,right = i+1;            int leftVal = cur,rightVal = cur;            //左移            while(left >= 0){                if(nums[left] < leftVal){                    count++;                    leftVal = nums[left];                }                left--;            }            //右移            while(right < nums.length){                if(nums[right] > rightVal){                    count++;                    rightVal = nums[right];                }                right++;            }            if(count > max)                max = count;        }        return max+1;    }}

//O(nlogn)public class Solution {    public int lengthOfLIS(int[] nums) {        if(nums.length <= 1)            return nums.length;        return binarySearch(nums,0,nums.length-1);    }    private int binarySearch(int[] nums,int left,int right){        if(left == right)            return 1;        if(right - left == 1){            if(nums[right] > nums[left])                return 2;            else if(nums[right] == nums[left])                return 1;            else                return 0;        }        int mid = (left+right)/2;        int leftlen = binarySearch(nums,left,mid);        int rightlen = binarySearch(nums,mid,right);        int count = 1;        int leftVal = nums[mid],rightVal = nums[mid];        int leftIndex = mid-1;        int rightIndex = mid+1;        while(leftIndex >= 0){            if(nums[leftIndex] < leftVal){                count++;                leftVal = nums[leftIndex];            }                leftIndex--;        }        while(rightIndex < nums.length){            if(nums[rightIndex] > rightVal){                count++;                rightVal = nums[rightIndex];            }            rightIndex++;        }        int tmpMax = Math.max(leftlen,rightlen);        return Math.max(tmpMax,count);    }}

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