283. Move Zeros
Given an array nums, 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, given nums = [0, 1, 0, 3, 12], after calling your function, nums should be [1, 3, 12, 0, 0].
Note:
You must do this in-place without making a copy of the array.
Minimize the total number of operations.
一根指针指向第一个非0的位置，第二根第一个0(如果不是0，自己跟自己换）
public void moveZeroes(int[] nums) {
if (nums == null || nums.length < 2) {
return;
}
int j = 0;
for (int i = 0; i < nums.length; i++) {
if (nums[i] != 0) {
// swap
int temp = nums[i];
nums[i] = nums[j];
nums[j] = temp;
j++;
}
}
}
360. Sort Transformed Array
Given a sorted array of integers nums and integer values a, b and c. Apply a quadratic function of the form f(x) = ax2 + bx + c to each element x in the array.
The returned array must be in sorted order.
Expected time complexity: O(n)
Example:
nums = [-4, -2, 2, 4], a = 1, b = 3, c = 5,
Result: [3, 9, 15, 33]
nums = [-4, -2, 2, 4], a = -1, b = 3, c = 5
Result: [-23, -5, 1, 7]
关键是在O(n)内返回sorted过的结果，就要利用a的符号来判断。假设a大于0，说明两端的值比中间大，要想得到一个从小到大的数组就要把值从后往前填。因为给的数组也是排过序的，一根指针指头一个指尾，往中间凑。想清楚while里的是小于等于号。
public int[] sortTransformedArray(int[] nums, int a, int b, int c) {
// if a = 0, b's sign doesn't matter
int left = 0, right = nums.length - 1;
int[] res = new int[nums.length];
int idx = (a >= 0) ? nums.length - 1 : 0;
while (left <= right) {
if (a >= 0) {
if (cal(a, b, c, nums[left]) >= cal(a, b, c, nums[right])) {
res[idx--] = cal(a, b, c, nums[left]);
left++;
} else {
res[idx--] = cal(a, b, c, nums[right]);
right--;
}
} else {
if (cal(a, b, c, nums[left]) >= cal(a, b, c, nums[right])) {
res[idx++] = cal(a, b, c, nums[right]);
right--;
} else {
res[idx++] = cal(a, b, c, nums[left]);
left++;
}
}
}
return res;
}