@[TOC]leetcode
Given an array, rotate the array to the right by k steps, where k is non-negative.
Example 1:
Input: [1,2,3,4,5,6,7] and k = 3
Output: [5,6,7,1,2,3,4]
Explanation:
rotate 1 steps to the right: [7,1,2,3,4,5,6]
rotate 2 steps to the right: [6,7,1,2,3,4,5]
rotate 3 steps to the right: [5,6,7,1,2,3,4]
原始思路
class Solution {
public:
void rotate(vector<int>& nums, int k) {
while(k){
int temp=nums[nums.size()-1];
nums.insert(nums.begin(),temp);
nums.pop_back();
k--;
}
}
};
STL思路
class Solution {
public void rotate(int[] nums, int k) {
// 旋转即是元素顺序轮转的意思
if (nums.length < 2 || k < 1 || k % nums.length == 0) {
return;
}
// 处理 k 大于 数组长度的情况
if (k > nums.length) {
k = k % nums.length;
}
// 对前 n - k 个元素 [1,2,3,4] 进行逆转后得到 [4,3,2,1]
reverse(nums, 0, nums.length - 1 - k);
// 对后k个元素 [5,6,7] 进行逆转后得到 [7,6,5]
reverse(nums, nums.length - k, nums.length -1);
// 将前后元素 [4,3,2,1,7,6,5] 逆转得到:[5,6,7,1,2,3,4]
reverse(nums, 0, nums.length - 1);
}
// 逆转数组指定区间内的元素,比如 [1,2,3,4] 逆转后变成 [4,3,2,1]
private void reverse(int[] nums, int start, int end) {
while (start < end) {
int temp = nums[start];
nums[start++] = nums[end];
nums[end--] = temp;
}
}
}
容易误解:top还是top,按照原来出栈入栈的规则即可,只不过是记录了最小值。
class MinStack {
public: /** initialize your data structure here. */
MinStack() {
min_val = INT_MAX;}
void push(int x){
if (x <= min_val) {
st.push(min_val);
min_val = x; }
st.push(x);
}
void pop() {
int t = st.top();
st.pop();
if (t == min_val) {
min_val = st.top();
st.pop();
}
}
int top() {
return st.top(); }
int getMin() {
return min_val;
}
private:
int min_val;
stack<int> st;
};