算法：接雨水
題目：?
給定?n個非負整數表示每個寬度為 1 的柱子的高度圖，計算按此排列的柱子，下雨之后能接多少雨水。
上面是由數組 [0,1,0,2,1,0,1,3,2,1,2,1] 表示的高度圖，在這種情況下，可以接 6 個單位的雨水（藍色部分表示雨水）
//暴力 int trap(vector<int>& height) {int ans = 0;int size = height.size();for (int i = 1; i < size - 1; i++) {int max_left = 0, max_right = 0;//左邊最大值for (int j = i; j >= 0; j--) { //Search the left part for max bar sizemax_left = max(max_left, height[j]);}//右邊最大值for (int j = i; j < size; j++) { //Search the right part for max bar sizemax_right = max(max_right, height[j]);}ans += min(max_left, max_right) - height[i];}return ans; }//動態編程 int trap(vector<int>& height) {if (height == null)return 0;int ans = 0;int size = height.size();vector<int> left_max(size), right_max(size);//把左邊最大值和右邊最大值不放在循環內部求解left_max[0] = height[0];for (int i = 1; i < size; i++) {left_max[i] = max(height[i], left_max[i - 1]);}right_max[size - 1] = height[size - 1];for (int i = size - 2; i >= 0; i--) {right_max[i] = max(height[i], right_max[i + 1]);}for (int i = 1; i < size - 1; i++) {ans += min(left_max[i], right_max[i]) - height[i];}return ans; }//棧的應用,將數組元素代入就能明白過程 int trap(vector<int>& height) {int ans = 0, current = 0;stack<int> st;while (current < height.size()) {//當前元素大于棧頂元素while (!st.empty() && height[current] > height[st.top()]) {int top = st.top();st.pop();if (st.empty())break;int distance = current - st.top() - 1;int bounded_height = min(height[current], height[st.top()]) - height[top];ans += distance * bounded_height;}//入棧st.push(current++);}return ans; }//代入值 int trap(vector<int>& height) {int left = 0, right = height.size() - 1;int ans = 0;int left_max = 0, right_max = 0;while (left < right) {if (height[left] < height[right]) {height[left] >= left_max ? (left_max = height[left]) : ans += (left_max - height[left]);++left;}else {height[right] >= right_max ? (right_max = height[right]) : ans += (right_max - height[right]);--right;}}return ans; }?
總結
