題目

https://leetcode.com/problems/count-submatrices-with-all-ones/

題解

本題與 leetcode 84. Largest Rectangle in Histogram | 84. 柱狀圖中最大的矩形（單調棧） 思路相似，借鑒了原來的代碼。

以及類似問題：leetcode 85. Maximal Rectangle | 85. 最大矩形（單調棧）

class Solution {public int numSubmat(int[][] mat) {if (mat.length == 0) return 0;int M = mat.length;int N = mat[0].length;int[][] heights = new int[M][N]; // 上方有多少個連續的1for (int i = 0; i < N; i++) {heights[0][i] = mat[0][i] - 0;for (int j = 1; j < M; j++) {if (mat[j][i] == 1) heights[j][i] = heights[j - 1][i] + 1;else heights[j][i] = 0;}}int result = 0;for (int i = 0; i < M; i++) {result += countRectangle(heights[i]);}return result;}// Reference:// leetcode 84. Largest Rectangle in Histogram// leetcode 85. Maximal Rectanglepublic int countRectangle(int[] heights) {int L = heights.length;// 找左邊第一個小于h[i]的數// 從右向左遍歷，維護單調不減棧，小數h[i]不斷將大數h[j]彈出，則h[i]左邊第一個小于h[i]的數為h[j]Stack<Integer> valueStack = new Stack<>();Stack<Integer> indexStack = new Stack<>();int[] leftIndex = new int[L]; // i左邊第一個小于i的數的下標Arrays.fill(leftIndex, -1);for (int i = L - 1; i >= 0; i--) {while (!valueStack.isEmpty() && valueStack.peek() > heights[i]) {leftIndex[indexStack.pop()] = i;valueStack.pop();}valueStack.push(heights[i]);indexStack.push(i);}// 找右邊第一個小于h[i]的數// 從左向右遍歷，維護單調不減棧valueStack = new Stack<>();indexStack = new Stack<>();int[] rightIndex = new int[L]; // i右邊第一個小于i的數的下標Arrays.fill(rightIndex, L);for (int i = 0; i < L; i++) {while (!valueStack.isEmpty() && valueStack.peek() > heights[i]) {rightIndex[indexStack.pop()] = i;valueStack.pop();}valueStack.push(heights[i]);indexStack.push(i);}// 對于每個h[i]，以當前高度分別向左右擴張，計算整個大區域的長方形個數int totalRectangle = 0;boolean[][] seen = new boolean[L + 2][L + 2]; // from,tofor (int i = 0; i < L; i++) {if (seen[leftIndex[i] + 1][rightIndex[i] + 1]) continue; // 相同高度只需計算一次else seen[leftIndex[i] + 1][rightIndex[i] + 1] = true;int leftHeight = leftIndex[i] >= 0 ? heights[leftIndex[i]] : 0; // 左邊第一個小于h[i]的高度int rightHeight = rightIndex[i] < L ? heights[rightIndex[i]] : 0; // 右邊第一個小于h[i]的高度int validHeight = heights[i] - Math.max(leftHeight, rightHeight); // 自由的高度int validWidth = rightIndex[i] - leftIndex[i] - 1; // 自由的寬度if (heights[i] > 0) totalRectangle += (validWidth * (validWidth + 1) / 2) * validHeight;}return totalRectangle;} }

總結

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