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【LeetCode & 劍指offer 刷題筆記】目錄（持續更新中...）

26. Remove Duplicates from Sorted Array

Given a sorted array nums, remove the duplicates in-place such that each element appear only once and return the new length. Do not allocate extra space for another array, you must do this by modifying the input array

in-place with O(1) extra memory. Example 1: Given nums = [1,1,2], Your function should return length = 2, with the first two elements of nums being 1 and 2 respectively. It doesn‘t matter what you leave beyond the returned length. Example 2: Given nums = [0,0,1,1,1,2,2,3,3,4], Your function should return length = 5, with the first five elements of nums being modified to 0, 1, 2, 3, and 4 respectively. It doesn‘t matter what values are set beyond the returned length. Clarification: Confused why the returned value is an integer but your answer is an array? Note that the input array is passed in by reference, which means modification to the input array will be known to the caller as well. Internally you can think of this: // nums is passed in by reference. (i.e., without making a copy) int len = removeDuplicates(nums); // any modification to nums in your function would be known by the caller. // using the length returned by your function, it prints the first len elements. for (int i = 0; i < len; i++) { print(nums[i]); } /* 問題：去除有序序列中的重復數字 方法一：雙指針法(覆蓋法) 當a[i] != a[index-1] 用a[i]覆蓋a[index] 相等時不覆蓋，不等時覆蓋，index代表了新數組的索引，i代表了舊數組索引,將無重復數依次移動到前面 如： 1 2 2(index) 2 3(i) 3 4 5 a[index] = a[i]; index++; O(n),O(1) */ class Solution { public: int removeDuplicates(vector<int>& a) { if(a.empty()) return 0; int index = 1; //初始時index和i指針均指向索引為1的位置 for(int i =1; i < a.size(); i++) //用i指針掃描數組 { if(a[i] != a[index-1]) //遇到不等數時，覆蓋a[index]，index++ { a[index] = a[i]; index++; } //index最後指向無重復序列末尾 } return index; //返回無重復序列的長度 } }; 80. Remove Duplicates from Sorted Array II 描述： Follow up for ”Remove Duplicates”: What if duplicates are allowed at most twice? For example, Given sorted array A = [1,1,1,2,2,3], Your function should return length = 5, and A is now [1,1,2,2,3] /* 覆蓋法 如： 1 2 2 2(index) 3(i) 3 4 5 a[index] = a[i]; index++; O(n),O(1) */ class Solution { public: int removeDuplicates(vector<int>& a) { int n = a.size(); if(n <= 2) return n; int index = 2; for(int i = 2; i<n; i++) { if(a[i] != a[index - 2]) { a[index] = a[i]; index++; } } return index; } };

【LeetCode & 劍指offer刷題】數組題6：26. Remove Duplicates from Sorted Array