Day 59 单调栈
503. 下一个更大元素II
class Solution {
public:vector<int> nextGreaterElements(vector<int>& nums) {int len = nums.size();nums.insert(nums.end(), nums.begin(), nums.end());vector<int> descStk, rst(len, -1);for (int i = 0; i < nums.size(); ++i){while (!descStk.empty() && nums[descStk.back()] < nums[i]){int idx = descStk.back() % len;if (rst[idx] == -1){rst[idx] = nums[i];}descStk.pop_back();}descStk.push_back(i);}return rst;}
};
其实也可以不扩充nums，而是在遍历的过程中模拟走了两边nums
class Solution {
public:vector<int> nextGreaterElements(vector<int>& nums) {int len = nums.size();vector<int> descStk, rst(len, -1);for (int i = 0; i < len * 2; ++i){int modI = i % len;while (!descStk.empty() && nums[descStk.back()] < nums[modI]){int idx = descStk.back() % len;if (rst[idx] == -1){rst[idx] = nums[modI];}descStk.pop_back();}descStk.push_back(modI);}return rst;}
};
42. 接雨水
三种方法都写了一遍
暴力方法
逐列计算
找到当前位置，左边最高的柱子的索引和右边最高的柱子的索引，两者中相对小的那个，将去当前位置的高度，就是水柱的高度，而宽度是1
超时了
class Solution {
public:int trap(vector<int>& height) {int sum = 0;for (int i = 1; i < height.size() - 1; i++){int leftTaller, rightTaller;leftTaller = rightTaller = height[i];for (int j = i - 1; j >= 0; --j){if (height[j] > leftTaller){leftTaller = height[j];}}for (int j = i + 1; j < height.size(); ++j){if (height[j] > rightTaller){rightTaller = height[j];}}sum += min(leftTaller, rightTaller) - height[i];}return sum;}
};
双指针
提前记录好左边最高的和右边最高的索引
class Solution {
public:int trap(vector<int>& height) {vector<int> maxLeft(height.size()), maxRight(height.size());maxLeft[0] = height[0];for (int i = 1; i < height.size(); ++i){maxLeft[i] = max(height[i], maxLeft[i - 1]);}maxRight.back() = height.back(); for (int i = height.size() - 2; i >= 0; --i){maxRight[i] = max(height[i], maxRight[i + 1]);}int sum = 0; for (int i = 1; i < height.size() - 1; i++){sum += min(maxLeft[i], maxRight[i]) - height[i];}return sum;}
};
暴力方法和双指针都是按照列的方式进行计算的
单调栈
而单调栈的方式是按照行的方式进行计算的，在思考的时候要记得这一点区别
class Solution {
public:int trap(vector<int>& height) {int sum = 0;stack<int> stk;stk.push(0);for (int i = 0; i < height.size(); ++i){if (height[i] < height[stk.top()]){stk.push(i);}else if (height[i] == height[stk.top()]){stk.pop();stk.push(i);}else{while (!stk.empty() && height[i] > height[stk.top()]){int idx = stk.top();stk.pop();if (!stk.empty()){int w = i - stk.top() - 1;int h = min(height[i], height[stk.top()]) - height[idx];sum += w * h;}}stk.push(i);}}return sum;}
};
)
