分析

題目

Given an array nums and a value val, 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 by modifying the input array in-place with O(1) extra memory. The order of elements can be changed. It doesn’t matter what you leave beyond the new length.

Example 1:

Given nums = [3,2,2,3], val = 3,

Your function should return length = 2, with the first two elements of nums being 2.

It doesn’t matter what you leave beyond the returned length.

Example 2:

Given nums = [0,1,2,2,3,0,4,2], val = 2,

Your function should return length = 5, with the first five elements of nums containing 0, 1, 3, 0, and 4.

Note that the order of those five elements can be arbitrary.

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 = removeElement(nums, val); // 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]); }

解答

package LeetCode;

public class L27_RemoveElement {
    public int removeElement(int[] nums, int val) {
        /*
        返回移除val後的陣列長度
        同時需要對nums進行修改，使得nums前len個數不包含val值
        前len個數的順序不做要求
        考慮類似快排的交換方式
         */
        int len=nums.length;
        int start=0,end=len-1;
        while(end>=start){
            //從後往前比較
            while(end>=start&&nums[end]==val)  //如果當前值等於val，比較下一個，直到有不等於val的交換位置，然後又從前往後比較
            {
                end--;
                len--;
            }
            //從前往後比較
            while(end>=start&&nums[start]!=val)//如果沒有等於val的，比較下一個，直到有比等於val的交換位置
                start++;
            if(end>=start){
                int temp = nums[start];
                nums[start] = nums[end];
                nums[end] = temp;
            }
            //一次迴圈交換一對數值的位置
        }
        // 比較結束，nums即基本有序。左邊的值都不等於val，右邊的值等於val
        return len;
    }
    public static void main(String[] args){
        int[] nums={1};
        int val=1;
        L27_RemoveElement re=new L27_RemoveElement();
        int len = re.removeElement(nums, val);
        // 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++) {
            System.out.println(nums[i]);
        }
    }
}