算法训练营第六十天打卡|84.柱状图中最大的矩形

Leetcode84.柱状图中最大的矩形

文章链接：代码随想录
文章链接：84.柱状图中最大的矩形

思路：暴力双指针，超时

class Solution {
public:int largestRectangleArea(vector<int>& heights) {int result = 0;for (int i = 0; i < heights.size(); i++){int l = i;int r = i;for (; l >= 0; l--){if (heights[l] < heights[i]) break;}for (; r < heights.size(); r++){if (heights[r] < heights[i]) break;}int w = r - l - 1;int h = heights[i];result = max(result, w * h);}return result;}
};

优化1（自己优化），依然超时，本质上还是没优化，时间复杂度一样

class Solution {
public:int largestRectangleArea(vector<int>& heights) {int result = 0;vector<int> nexl(heights.size(), -1);vector<int> nexr(heights.size(), heights.size());for (int i = 0; i < heights.size(); i++){for (int j = i - 1; j >= 0; j--){if (heights[j] < heights[i]){nexl[i] = j;break;}}}for (int i = heights.size() - 1; i >= 0; i--){for (int j = i + 1; j < heights.size(); j++){if (heights[j] < heights[i]){nexr[i] = j;break;}}}for (int i = 0; i < heights.size(); i++){int w = nexr[i] - nexl[i] - 1;int h = heights[i];result = max(result, w * h);}return result;}
};

优化了双指针，在优化过程中用上了之前记录的nexl、nexr中的信息，降低了时间复杂度

class Solution {
public:int largestRectangleArea(vector<int>& heights) {int result = 0;vector<int> nexl(heights.size(), -1);vector<int> nexr(heights.size(), heights.size());for (int i = 0; i < heights.size(); i++){int j = i - 1;while (j >= 0 && heights[j] >= heights[i]) j = nexl[j];nexl[i] = j;} for (int i = heights.size() - 1; i >= 0; i--){int j = i + 1;while (j < heights.size() && heights[j] >= heights[i]) j = nexr[j];nexr[i] = j;}for (int i = 0; i < heights.size(); i++){int w = nexr[i] - nexl[i] - 1;int h = heights[i];result = max(result, w * h);}return result;}
};

单调栈，和接雨水正好相反，是找元素两边第一个比本身小的，数组前后需要各插入一个0，避免递增或递减序列无法正确计算

class Solution {
public:int largestRectangleArea(vector<int>& heights) {int result = 0;heights.insert(heights.begin(), 0);heights.push_back(0);//heights.insert(heights.end(), 0);stack<int> st;st.push(0);for (int i = 1; i < heights.size(); i++){if (heights[i] > heights[st.top()]) st.push(i);else if (heights[i] == heights[st.top()]) {st.pop();st.push(i);}else{while(!st.empty() && heights[i] < heights[st.top()]){int mid = st.top();st.pop();if (!st.empty()){int w = i - st.top() - 1;int h = heights[mid];result = max(result, h * w);}}st.push(i);}}return result;}
};