hot100之子串

和为K的子数组(560)

先看代码

class Solution {
    public int subarraySum(int[] nums, int k) {
        int res = 0;
        int preSum = 0;
        Map<Integer, Integer> cnt = new HashMap<>(nums.length);
        for (int num : nums){
            cnt.merge(preSum, 1, Integer::sum);
            preSum += num;
            res += cnt.getOrDefault(**preSum - k**, 0);
        }
        return res;
    }
}

• 分析

cnt用来记录相同前缀和的前缀数量

一次循环一边更新相同前缀和的前缀数量, 一边将吻合的数据放入res

preSum - k 是前缀和的关键

• 感悟

只有”单调”的数据集才适合用滑动窗口, 非”单调”的前缀和才好做 , 否则每次移动窗口都带来混乱

“单调”指的是 任意 num>0且单调递增 OR 任意 num < 0 且单调递减

不符合单调

符合单调

滑动窗口最大值(560)

先看代码

class Solution {
    public int[] maxSlidingWindow(int[] nums, int k) {
        int n = nums.length;
        int[] res = new int[n-k+1];
        Deque<Integer> queue = new ArrayDeque<>();
        for (int i = 0; i < n; i++){
            while (!queue.isEmpty() && nums[queue.getLast()] <= nums){
                queue.removeLast();
            }
            queue.addLast(i);

            if (i-k >= queue.getFirst()){
                queue.removeFirst();
            }
            if(i-k+1 >= 0) res[i-k+1] = nums[queue.getFirst()];
        }
        return res;
    }
}

class Solution {
    public String minWindow(String s, String t) {
        int needCnt = 0;
        int[] need = new int[128];
        for (char c : t.toCharArray()){
            if (need[c] == 0){
                needCnt++;
            }need[c]++;
        }
        int n = s.length();
        int resLef = -1;
        int resRig = n;

        int lef = 0;
        for (int rig = 0; rig < n; rig++){
            char cR = s.charAt(rig);
            need[cR]--;
            if (need[cR] == 0){
                needCnt--;
            }
            while (needCnt == 0){
                if(resRig-resLef > rig-lef){
                    resLef = lef;
                    resRig = rig; 
                }
                char cL = s.charAt(lef);
                if (need[cL] == 0) needCnt++;
                need[cL]++;
                lef++;
            } 
        }

        return resLef < 0 ? "" : s.substring(resLef, resRig+1);
    }
}