【当前排名67359】挑战leetcode进入前1w名

李恒道

https://leetcode.cn/problems/count-submatrices-with-all-ones/submissions/587906655/
这题dp真的很有意思
简单来说将第一个计算只有一行的，第二次计算两行的，第三次三行的

var numSubmat = function (mat) {
  let total = 0;
  for (let col = 0; col < mat.length; col++) {
    for (let index = col; index < mat.length; index++) {
      let ans = 0;
      for (let indey = 0; indey < mat[0].length; indey++) {
        const num = mat[index][indey];
        if (num == 0) {
          ans = 0;
        } else {
          ans = ans + 1;
        }
        total += ans;
      }
    }
    for (let index = mat.length - 1; index > 0; index--) {
      for (let indey = 0; indey < mat[0].length; indey++) {
        mat[index][indey] &= mat[index - 1][indey];
      }
    }
  }

  return total;
};

李恒道

https://leetcode.cn/problems/string-to-integer-atoi/submissions/587908846/
没去凹边界极大值情况
讲道理不太喜欢那种模板题

var myAtoi = function (s) {
  let result = 0;
  let hasPrefix = "";
  for (let index = 0; index < s.length; index++) {
    const charCode = s[index].charCodeAt();
    const char = s[index];
    if (char == " ") {
      if(hasPrefix!==""){
        break;
      }
      continue;
    }
    if (char == "+" || char == "-") {
      if (hasPrefix === "") {
        hasPrefix = char == "+" ? 1 : -1;
      } else {
        break;
      }
      continue;
    }
    if (charCode >= 48 && charCode <= 57) {
      result = result * 10 + (charCode - 48);
      if (hasPrefix === "") {
        hasPrefix = 1;
      }
    } else {
      break;
    }
    if(result>(2**31)&&hasPrefix==-1){
      return (2**31)*-1
    }
    if(result>(2**31-1)&&hasPrefix==1){
      return (2**31-1)
    }
  }
  return result * (hasPrefix === "" ? 1 : hasPrefix);
};

李恒道

https://leetcode.cn/problems/sum-of-subarray-minimums/submissions/588105406/
脑子没想明白
手凹过去了
我草

var sumSubarrayMins = function (arr) {
    const left = new Array(arr.length).fill(-1)
    const right = new Array(arr.length).fill(-1)
    const st = [];
    for (let i = 0; i < arr.length; i++) {
        const x = arr[i];
        while (st.length && x <= arr[st[st.length - 1]]) {
            st.pop();
        }
        if (st.length) {
            left[i] = st[st.length - 1];
        }
        st.push(i);
    }
    st.length = 0;
    for (let i = arr.length - 1; i >= 0; i--) {
        const x = arr[i];
        while (st.length && x < arr[st[st.length - 1]]) {
            st.pop();
        }
        if (st.length) {
            right[i] = st[st.length - 1];
        }
        st.push(i);
    }
    let ans = BigInt(0);
    for (let i = 0; i < arr.length; i++) {
        let total = 1;
        const l = left[i];
        const r = right[i];
        let leftLen=0;
        if (l !== -1) {
            leftLen += i - l - 1
        } else {
            leftLen += i
        }
        let rightLen=0;
        if (r !== -1) {
            rightLen += r - i - 1
        } else {
            rightLen += arr.length - i - 1
        }
        if (leftLen!==0 && rightLen!==0) {
            total += leftLen * rightLen
        }
        total+=leftLen+rightLen
        ans +=( BigInt(total) * BigInt(arr[i]))% BigInt(10**9+7)
        ans =ans %BigInt(10**9+7)

    }
    return Number(ans);
}

李恒道

https://leetcode.cn/problems/surrounded-regions/submissions/588110103/?envType=study-plan-v2&envId=top-interview-150
一波秒

var solve = function (board) {
    const dfs = (x, y, target, replace) => {
        if (x < 0 || x >= board.length || y < 0 || y >= board[0].length) {
            return
        }
        if (board[x][y] == target) {
            board[x][y] = replace
            dfs(x + 1, y, target, replace)
            dfs(x - 1, y, target, replace)
            dfs(x, y + 1, target, replace);
            dfs(x, y - 1, target, replace)
        }
    }
    for (let index = 0; index < board[0].length; index++) {
        if (board[0][index] == "O") {
            dfs(0, index, "O", "Z")
        }
        if (board[board.length - 1][index] == "O") {
            dfs(board.length - 1, index, "O", "Z")
        }
    }
    for (let index = 0; index < board.length; index++) {
        if (board[index][0] == "O") {
            dfs(index, 0, "O", "Z")
        }
        if (board[index][board[0].length - 1] == "O") {
            dfs(index, board[0].length - 1, "O", "Z")
        }
    }
    for (let index = 0; index < board.length; index++) {
        for (let indey = 0; indey < board[0].length; indey++) {
            if (board[index][indey] == "O") {
                dfs(index, indey, "O", "X")
            }
        }
    }
    for (let index = 0; index < board.length; index++) {
        for (let indey = 0; indey < board[0].length; indey++) {
            if (board[index][indey] == "Z") {
                dfs(index, indey, "Z", "O")
            }
        }
    }

};

李恒道

https://leetcode.cn/problems/find-indices-of-stable-mountains/submissions/588113165/
简单题
过

var stableMountains = function(height, threshold) {
    let left=height[0];
    let ans=[]
    for (let index = 1; index < height.length; index++) {
        if(left>threshold){
            ans.push(index)
        }
        left=height[index]
    }
    return ans
};

李恒道

https://leetcode.cn/problems/game-of-life/submissions/588119767/?envType=study-plan-v2&envId=top-interview-150
位运算题
秒了

var gameOfLife = function (board) {
    const getNear = (x, y) => {
        const getPos = (x, y) => {
            if (x < 0 || x >= board.length || y < 0 || y > board[0].length) {
                return 0
            }
            return ((board[x][y] & 1) == 1) ? 1 : 0
        }
        return getPos(x + 1, y) + getPos(x - 1, y) + getPos(x, y + 1) + getPos(x, y - 1) +
            getPos(x - 1, y - 1) + getPos(x - 1, y + 1) + getPos(x + 1, y - 1) + getPos(x + 1, y + 1)
    }
    for (let index = 0; index < board.length; index++) {
        for (let indey = 0; indey < board[0].length; indey++) {
            let status = board[index][indey];
            const count = getNear(index, indey)
            if (status == 1) {
                //活细胞
                if (count == 2 || count == 3) {
                    status = status | (1 << 1)
                }
            } else {
                if (count == 3) {
                    status = status | (1 << 1)
                }
            }
            board[index][indey] = status
        }
    }
    for (let index = 0; index < board.length; index++) {
        for (let indey = 0; indey < board[0].length; indey++) {
            board[index][indey] = (board[index][indey] >> 1)
        }
    }
};

李恒道

https://leetcode.cn/problems/path-sum/submissions/588123625/?envType=study-plan-v2&envId=top-interview-150
简单题，秒了

var hasPathSum = function (root, targetSum) {
    if (root === null) {
        return false
    }
    const dfs = (node, val) => {
        val = val - node.val;
        if (node.left == null && node.right == null) {
            return val == 0
        }
        return (node.left && dfs(node.left, val)) || (node.right && dfs(node.right, val))||false
    }
    return dfs(root, targetSum)
};

李恒道

https://leetcode.cn/problems/minimum-length-of-anagram-concatenation/submissions/588290954/
这题写一半卡住了
边界没搞明白
mark一下

const check = (str, len) => {
    let arr1 = new Array(26).fill(0)
    for (let index = 0; index < str.length; index += len) {
        const arr2 = new Array(26).fill(0)
        for (let pos = index; pos < index + len; pos++) {
            arr2[str[pos].charCodeAt() - 97]++
        }
        const result = arr2.every((val, index) => arr1[index] == val)
        if(index==0||result==true){
            arr1=arr2
        }else{
            return false
        }
    }
    return true
}
var minAnagramLength = function (s) {
    for (let index = 1; index < s.length; index++) {
        if (s.length % (index) !== 0) {
            continue
        }
        if (check(s, index)) {
            return index
        }
    }
    return s.length
};

李恒道

https://leetcode.cn/problems/single-number-ii/submissions/588293666/?envType=study-plan-v2&envId=top-interview-150
秒了

var singleNumber = function (nums) {
    let num = 0
    for (let index = 0; index < 32; index++) {
        let total = 0;
        for (let indey = 0; indey < nums.length; indey++) {
            const num = nums[indey];
            total += ((num & (1 << index)) == 0) ? 0 : 1
        }
        num=num|((total%3)<<index)
    }
    return num
};

王一之

有排名了吗？