单调栈
求解一个集合中,每个元素左边离自己最近且比自己小的元素是?右边离自己最近且比自己小的元素是?
没有重复值:
public int[][] getNearLessNoRepeat(int[] arr){
int[][] res = new int[arr.length][2];
Stack<Integer> stack = new Stack<>();
for (int i = 0; i < res.length; i++) {
while(!stack.isEmpty() && arr[stack.peek()] > arr[i]) {
int j = stack.pop();
int leftLessIndex = stack.isEmpty() ? -1 : stack.peek();
res[j][0] = leftLessIndex;
res[j][1] = i;
}
stack.push(i);
}
while(!stack.isEmpty()) {
int j = stack.pop();
int leftLessIndex = stack.isEmpty() ? -1 : stack.peek();
res[j][0] = leftLessIndex;
res[j][1] = -1;
}
return res;
}
有重复值:
public int[][] getNearLess(int[] arr) {
int[][] res = new int[arr.length][2];
Stack<List<Integer>> stack = new Stack<>();
for (int i = 0; i < arr.length; i++) { // i -> arr[i] 进栈
while (!stack.isEmpty() && arr[stack.peek().get(0)] > arr[i]) {
List<Integer> popIs = stack.pop();
int leftLessIndex = stack.isEmpty() ? -1 : stack.peek().get(stack.peek().size() - 1);
for (Integer popi : popIs) {
res[popi][0] = leftLessIndex;
res[popi][1] = i;
}
}
if (!stack.isEmpty() && arr[stack.peek().get(0)] == arr[i]) {
stack.peek().add(Integer.valueOf(i));
} else {
ArrayList<Integer> list = new ArrayList<>();
list.add(i);
stack.push(list);
}
}
while (!stack.isEmpty()) {
List<Integer> popIs = stack.pop();
int leftLessIndex = stack.isEmpty() ? -1 : stack.peek().get(stack.peek().size() - 1);
for (Integer popi : popIs) {
res[popi][0] = leftLessIndex;
res[popi][1] = -1;
}
}
return res;
}
给定一个只包含正数的数组arr,arr中任何一个子数组sub,一定都可以算出(sub累加和 )* (sub中的最小值)是什么,那么所有子数组中,这个值最大是多少?
暴力解:
public static int max1(int[] arr) {
int max = Integer.MIN_VALUE;
for (int i = 0; i < arr.length; i++) {
for (int j = i; j < arr.length; j++) {
int minNum = Integer.MAX_VALUE;
int sum = 0;
for (int k = i; k <= j; k++) {
sum += arr[k];
minNum = Math.min(minNum, arr[k]);
}
max = Math.max(max, minNum * sum);
}
}
return max;
}
前缀和+单调栈
public static int max2(int[] arr) {
int size = arr.length;
int[] sums = new int[size];
sums[0] = arr[0];
for (int i = 1; i < size; i++) {
sums[i] = sums[i - 1] + arr[i];
}
int max = Integer.MIN_VALUE;
Stack<Integer> stack = new Stack<Integer>();
for (int i = 0; i < size; i++) {
while (!stack.isEmpty() && arr[stack.peek()] >= arr[i]) {
int j = stack.pop();
max = Math.max(max, (stack.isEmpty() ? sums[i - 1] : (sums[i - 1] - sums[stack.peek()])) * arr[j]);
}
stack.push(i);
}
while (!stack.isEmpty()) {
int j = stack.pop();
max = Math.max(max, (stack.isEmpty() ? sums[size - 1] : (sums[size - 1] - sums[stack.peek()])) * arr[j]);
}
return max;
}
84. 柱状图中最大的矩形
给定 n 个非负整数,用来表示柱状图中各个柱子的高度。每个柱子彼此相邻,且宽度为 1 。
求在该柱状图中,能够勾勒出来的矩形的最大面积。
示例 1:
输入:heights = [2,1,5,6,2,3]
输出:10
解释:最大的矩形为图中红色区域,面积为 10
示例 2:
输入: heights = [2,4]
输出: 4
public int largestRectangleArea(int[] h) {
if(h == null || h.length == 0) {
return 0;
}
int maxArea = 0;
Stack<Integer> stack = new Stack<>();
for (int i = 0; i < h.length; i++) {
while(!stack.isEmpty() && h[i] <= h[stack.peek()]) {
int j = stack.pop();
int k = stack.isEmpty() ? -1 : stack.peek();
int curArea = (i-k-1) * h[j];
maxArea = Math.max(maxArea, curArea);
}
stack.push(i);
}
while (!stack.isEmpty()) {
// 长
int j = stack.pop();
// 宽
int k = stack.isEmpty() ? -1 : stack.peek();
int curArea = (h.length -k-1) * h[j];
maxArea = Math.max(maxArea, curArea);
}
return maxArea;
}
}
每个人都有潜在的能量,只是很容易被习惯所掩盖,被时间所迷离,被惰性所消磨~