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[LintCode/LeetCode] Remove Element [Two Pointers]

EdwardUp / 995人閱讀

摘要：雙指針頭指針等于指定元素的時(shí)候，用尾指針的值替換的值否則頭指針繼續(xù)向后走。最后返回，就是所有非元素的數(shù)量。

Problem

Given an array and a value, remove all occurrences of that value in place and return the new length.

The order of elements can be changed, and the elements after the new length don"t matter.

Example

Given an array [0,4,4,0,0,2,4,4], value=4

return 4 and front four elements of the array is [0,0,0,2]

Note

雙指針I(yè)：頭指針i等于指定元素elem的時(shí)候，用尾指針j的值替換i的值（A[i] = A[--j]）；否則頭指針i繼續(xù)向后走。
雙指針I(yè)I：i和j都作為頭指針，當(dāng)i的值不是指定元素elem的時(shí)候，將A[i]復(fù)制到j(luò)的位置；否則i繼續(xù)向后走。最后返回j，就是所有非elem元素的數(shù)量。

Solution

1. 雙指針I(yè)

public class Solution {
    public int removeElement(int[] A, int elem) {
        int i = 0, j = A.length;
        while (i < j) {
            if (A[i] == elem) {
                A[i] = A[--j];
            }
            else i++;
        }
        return j;
    }
}

2. 雙指針I(yè)I

public class Solution {
    public int removeElement(int[] A, int elem) {
        int i = 0, j = 0;
        while (i < A.length) {
            if (A[i] != elem) {
                A[j++] = A[i];
            }
            i++;
        }
        return j;
    }
}

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