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add n*O(logn) solution (cpp)

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	`1`	`+class Solution {`
	`2`	`+// dp[i] 表示当前长度为i+1的最大升序左侧子序列的最小末尾值`
	`3`	`+// 可以知道dp[]一定是递增的,每多加一个元素就可以用折半查找更新dp`
	`4`	`+public:`
	`5`	`+ int lengthOfLIS(vector<int>& nums) {`
	`6`	`+ if (nums.size() == 0)`
	`7`	`+ return 0 ;`
	`8`	`+ int dp[nums.size()] = {nums[0], };`
	`9`	`+ int maxLen = 1 ; // 刚开始找到的最大左侧子串是第一个元素,长度为1`
	`10`	`+ for (int i = 1; i < nums.size(); ++i)`
	`11`	`+ {`
	`12`	`+ int l = 0, r = maxLen ;`
	`13`	`+ while (l < r)`
	`14`	`+ {`
	`15`	`+ const int mid = l + (r-l)/2 ;`
	`16`	`+ if (dp[mid] < nums[i])`
	`17`	`+ l = mid+1 ;`
	`18`	`+ else`
	`19`	`+ r = mid ;`
	`20`	`+ }`
	`21`	`+`
	`22`	`+ // 长度为l+1的最大上升子序列的最小末尾值可以更新为该元素了`
	`23`	`+ dp[l] = nums[i] ;`
	`24`	`+`
	`25`	`+ // 如果当前元素比dp中所有元素都大,最大长度就可以+1,并以该元素结尾`
	`26`	`+ if (l == maxLen)`
	`27`	`+ ++maxLen ;`
	`28`	`+ }`
	`29`	`+ return maxLen ;`
	`30`	`+ }`
	`31`	`+};`

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