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Merged
codedrinker merged 2 commits into from
May 29, 2019
Merged

wangyuhuich:Move Zeroes #1

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wangyuhuich:Move Zeroes
wangyuhuich May 29, 2019
af388a1
alter algorithm
wangyuhuich May 29, 2019
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55 changes: 55 additions & 0 deletions src/main/java/MoveZeroes/wangyuhuich/MoveZeroes.java
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Original file line number Diff line number Diff line change
@@ -0,0 +1,55 @@
package MoveZeroes.wangyuhuich;

/**
* 给定一个数组 nums，编写一个函数将所有 0 移动到数组的末尾，同时保持非零元素的相对顺序。
* <p>
* 示例:
* <p>
* 输入: [0,1,0,3,12]
* 输出: [1,3,12,0,0]
* 说明:
* <p>
* 必须在原数组上操作，不能拷贝额外的数组。
* 尽量减少操作次数。
*/

/**
* 解题思路:类似冒泡,数字0 上浮
*/


public class MoveZeroes {
/**
* 执行用时 : 1 ms, 在Move Zeroes的Java提交中击败了97.69% 的用户
* 内存消耗 : 39.9 MB, 在Move Zeroes的Java提交中击败了72.80% 的用户
*/
public void moveZeroes(int[] nums) {
int zeroCount = 0;
int notZeroPosition = 0;
for (int i = 0; i < nums.length; i++) {
if(nums[i]==0){
zeroCount+=1;
}else{
nums[notZeroPosition++] = nums[i];
}
}
for (int i = notZeroPosition,j = 0; j < zeroCount; i++,j++) {
nums[i] = 0;
}
}

/**
* 执行用时 : 122 ms, 在Move Zeroes的Java提交中击败了5.07% 的用户
* 内存消耗 : 40.6 MB, 在Move Zeroes的Java提交中击败了54.19% 的用户
*/
// @Deprecated
// public void moveZeroes(int[] nums) {
// for (int i = 0; i < nums.length; i++) {
// for (int j = 0; j < nums.length-i-1; j++) {
// if (nums[j] == 0 && nums[j + 1] != 0) {
// nums[j] = nums[j + 1] + (nums[j + 1] = nums[j]) - nums[j];
// }
// }
// }
// }
}