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Given an unsorted array of integers, find the length of longest increasing subsequence.
For example,
Given [10, 9, 2, 5, 3, 7, 101, 18],
The longest increasing subsequence is [2, 3, 7, 101], therefore the length is 4. Note that there may be more than one LIS combination, it is only necessary for you to return the length.
Your algorithm should run in O(n2) complexity.
Follow up: Could you improve it to O(n log n) time complexity?
https://leetcode.com/problems/longest-increasing-subsequence/
动态规划。
先是O(n2)的解法，dp[i]是以nums[i]结尾的最长子串的长度。
1 /** 2 * @param {number[]} nums 3 * @return {number} 4 */ 5 var lengthOfLIS = function(nums) { 6 var max = 0, dp = [], i, j, curMax; 7 for(i = 0; i < nums.length; i++){ 8 curMax = 0; 9 for(j = 0; j < i; j++){ 10 if(nums[i] > nums[j] && dp[j] > curMax){ 11 curMax = dp[j]; 12 } 13 } 14 curMax++; 15 if(curMax > max){ 16 max = curMax; 17 } 18 dp[i] = curMax; 19 } 20 return max; 21 };
O(logn)的解法，dp[i]是以dp[i]结尾的长度为i的子串。
比如[1,5,4,6]
1. dp = [1]
2. dp = [1,5]，长度为2的子串最小以5结尾
3. dp = [1,4]，长度为2的子串最小以4结尾
4. dp = [1,4,6]，长度为3的子串最小以6结尾
结果就是dp的长度3。
1 /** 2 * @param {number[]} nums 3 * @return {number} 4 */ 5 var lengthOfLIS = function(nums) { 6 var dp = [], i, j, len; 7 for(i = 0; i < nums.length; i++){ 8 len = dp.length; 9 if(len === 0 || dp[len - 1] < nums[i]){ 10 dp[len] = nums[i]; 11 }else{ 12 dp[bSearch(nums[i], 0, len - 1)] = nums[i]; 13 } 14 15 } 16 return dp.length; 17 18 function bSearch(num, i, j){ 19 if(i === j){ 20 return i; 21 } 22 var middle = parseInt((i + j) / 2); 23 if(dp[middle] < num){ 24 return bSearch(num, middle + 1, j); 25 } 26 return bSearch(num, i, middle); 27 } 28 };
[LeetCode][JavaScript]Longest Increasing Subsequence
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原文地址：http://www.cnblogs.com/Liok3187/p/4967239.html
