[LeetCode] Move Zeroes - 整数数组处理问题
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目录:
1.Move Zeroes - 数组0移到末尾 [顺序交换]
2.
题目概述:
Given an arraynums, write a function to move all 0's to the end of it while maintaining the relative order of the non-zero elements.For example, givennums = [0, 1, 0, 3, 12], after calling your function,nums should be[1, 3, 12, 0, 0].
Note:
1.You must do this in-place without making a copy of the array.
2.Minimize the total number of operations.
1.Move Zeroes - 数组0移到末尾 [顺序交换]
2.
一.Move Zeroes
题目概述:
Given an arraynums, write a function to move all 0's to the end of it while maintaining the relative order of the non-zero elements.For example, givennums = [0, 1, 0, 3, 12], after calling your function,nums should be[1, 3, 12, 0, 0].
Note:
1.You must do this in-place without making a copy of the array.
2.Minimize the total number of operations.
解题方法:
题意是把数组nums中0的元素后置,同时不能采用赋值数组。两种方法:
1.遇到是0的元素从数组最后向前存储并移位,遇到非0元素从前存储;
2.推荐:从前往后查找,不是0的元素前移,并计算0的个数,后面的全置0。
我的代码:
方法一:Runtime: 28 ms
void moveZeroes(int* nums, int numsSize) { int endNum; //从后计数0 int startNum; //从前计数非0 int temp; int i,j; i = 0; startNum = 0; endNum = 0; while( (i+endNum) < numsSize ) { if(nums[i]==0) { //依次前移 for(j=startNum; j<numsSize-endNum-1; j++) { //j少一个数 nums[j] = nums[j+1]; } nums[numsSize-endNum-1] = 0; endNum++; } else { nums[startNum] = nums[i]; startNum++; i++; } }}方法二:Runtime: 8 ms
void moveZeroes(int* nums, int numsSize) { int count; //计算0的个数 int i,j; int n; n = 0; count = 0; for(i=0; i<numsSize; i++) { if(nums[i]==0) { count++; } else { nums[n] = nums[i]; n++; } } //后置0 for(j=0; j<count; j++) { nums[n] = 0; n++; }}
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