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++;
}
}
