leetcode26 remove duplicate

題目要求：輸入一個數組和一個值，刪除數組中等於該值得元素。不容許分配新的內存空間（即不容許建立新的數組），容許數組中的元素的順序發生變化，只要該數組在返回長度前的值正確
例如：輸入nums = [3,2,2,3], val = 3，程序返回2，且nums數組的前兩個值均爲2

使用一個指針
時間複雜度O(n), 空間複雜度O(1)

/**
 * @author rale
 * Given an array and a value, remove all instances of that value in place and return the new length.
 * Do not allocate extra space for another array, you must do this in place with constant memory.
 * The order of elements can be changed. It doesn't matter what you leave beyond the new length.
 * Example:
 * Given input array nums = [3,2,2,3], val = 3
 * Your function should return length = 2, with the first two elements of nums being 2.
 */
public class RemoveElement {
    
    public int removeElement(int[] nums, int val) {
        int index = 0;
        for(int i = 0 ; i<nums.length ; i++){
            if(nums[i] != val){
                nums[index]=nums[i];
                index++;
            }
        }
        return index;
    }
}

使用兩個指針
時間複雜度O(n) 空間複雜度O(1)

/**
 * @author rale
 * Given an array and a value, remove all instances of that value in place and return the new length.
 * Do not allocate extra space for another array, you must do this in place with constant memory.
 * The order of elements can be changed. It doesn't matter what you leave beyond the new length.
 * Example:
 * Given input array nums = [3,2,2,3], val = 3
 * Your function should return length = 2, with the first two elements of nums being 2.
 */
public class RemoveElement {
    
    public int removeElement(int[] nums, int val) {
        if(nums==null || nums.length==0){
            return 0;
        }
        int leftPointer = 0;
        int rightPointer = nums.length-1;
        while(leftPointer<=rightPointer){
            if(nums[rightPointer]==val){
                rightPointer--;
                continue;
            }
            if(nums[leftPointer]==val){
                nums[leftPointer] = nums[rightPointer];
                rightPointer--;
            }
            leftPointer++;
        }
        return rightPointer+1;
    }
}

leetcode的測試用例得出的性能分析發現，使用一個指針比使用兩個指針的速度更快。可是在數組的容量很是大且數組中該數字出現頻率不高的狀況下，使用兩個指針能夠明顯減小程序遍歷數組的時間。
leetcode上也給出了參考答案

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