Commit 8806412

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Create Find First and Last Position of Element in Sorted Array

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	`1`	`+想法:用 binary search 找出第一個出現的位置跟最後一個出現的位置即可,沒有找到則輸出 {-1 , -1}`
	`2`	`+第一個出現的位置:相等時維護右界相同(代表第一個出現的元素是包含現在看的以前)`
	`3`	`+最後一個出現的位置:相等時維護左界相同(代表最後一個出現的元素是包含現在看的以後)`
	`4`	`+`
	`5`	`+Time Complexity : O(logn) for binary search two times`
	`6`	`+Space Complexity : O(1) for variables`
	`7`	`+`
	`8`	`+class Solution {`
	`9`	`+public:`
	`10`	`+ vector<int> searchRange(vector<int>& nums, int target) {`
	`11`	`+ // find the start position`
	`12`	`+ int l = 0 , r = nums.size() - 1 ;`
	`13`	`+ int start ;`
	`14`	`+ while (l < r) {`
	`15`	`+ int m = (l + r) / 2 ;`
	`16`	`+ if ( nums[m] > target )`
	`17`	`+ r = m - 1 ;`
	`18`	`+ else if ( nums[m] == target)`
	`19`	`+ r = m ;`
	`20`	`+ else`
	`21`	`+ l = m + 1 ;`
	`22`	`+ }`
	`23`	`+ if (nums.empty() \|\| nums[l] != target)`
	`24`	`+ return {-1 , -1} ;`
	`25`	`+ else`
	`26`	`+ start = l ;`
	`27`	`+`
	`28`	`+ // find the end position`
	`29`	`+ l = 0 , r = nums.size() - 1 ;`
	`30`	`+ while (l < r) {`
	`31`	`+ int m = (l + r + 1) / 2 ;`
	`32`	`+ if (nums[m] > target)`
	`33`	`+ r = m - 1 ;`
	`34`	`+ else if (nums[m] == target)`
	`35`	`+ l = m ;`
	`36`	`+ else`
	`37`	`+ l = m + 1 ;`
	`38`	`+ }`
	`39`	`+ return {start , l} ;`
	`40`	`+ }`
	`41`	`+};`

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