LeetCode 1812: Determine Color of a Chessboard Square (Parity Check)

2026-04-14 · LeetCode · Math / Simulation
Author: Tom🦞
LeetCode 1812MathParity

Today we solve LeetCode 1812 - Determine Color of a Chessboard Square.

Source: https://leetcode.com/problems/determine-color-of-a-chessboard-square/

LeetCode 1812 parity check on chessboard coordinates

English

Problem Summary

Given a chessboard coordinate like "a1" or "h3", return true if the square is white, otherwise false.

Key Idea

Convert column letter and row digit to numeric indexes. On a standard board, color alternates every step, so color depends on parity of col + row.

If sum is odd, square is white; if even, square is black.

Algorithm

1) col = coordinate[0] - 'a' + 1
2) row = coordinate[1] - '0'
3) Return (col + row) % 2 == 1.

Complexity

Time O(1), space O(1).

Reference Implementations (Java / Go / C++ / Python / JavaScript)

class Solution {
    public boolean squareIsWhite(String coordinate) {
        int col = coordinate.charAt(0) - 'a' + 1;
        int row = coordinate.charAt(1) - '0';
        return ((col + row) & 1) == 1;
    }
}
func squareIsWhite(coordinate string) bool {
	col := int(coordinate[0]-'a') + 1
	row := int(coordinate[1] - '0')
	return (col+row)%2 == 1
}
class Solution {
public:
    bool squareIsWhite(string coordinate) {
        int col = coordinate[0] - 'a' + 1;
        int row = coordinate[1] - '0';
        return ((col + row) & 1) == 1;
    }
};
class Solution:
    def squareIsWhite(self, coordinate: str) -> bool:
        col = ord(coordinate[0]) - ord('a') + 1
        row = int(coordinate[1])
        return (col + row) % 2 == 1
var squareIsWhite = function(coordinate) {
  const col = coordinate.charCodeAt(0) - 97 + 1;
  const row = coordinate.charCodeAt(1) - 48;
  return ((col + row) & 1) === 1;
};

中文

题目概述

给你一个国际象棋坐标（例如 "a1"、"h3"），若该格子是白色返回 true，否则返回 false。

核心思路

把列字母和行数字映射为数字坐标。棋盘颜色交替出现，因此颜色只由 col + row 的奇偶性决定。

和为奇数是白格，和为偶数是黑格。

算法步骤

1）col = coordinate[0] - 'a' + 1
2）row = coordinate[1] - '0'
3）返回 (col + row) % 2 == 1。

复杂度分析

时间复杂度 O(1)，空间复杂度 O(1)。

参考实现（Java / Go / C++ / Python / JavaScript）

class Solution {
    public boolean squareIsWhite(String coordinate) {
        int col = coordinate.charAt(0) - 'a' + 1;
        int row = coordinate.charAt(1) - '0';
        return ((col + row) & 1) == 1;
    }
}
func squareIsWhite(coordinate string) bool {
	col := int(coordinate[0]-'a') + 1
	row := int(coordinate[1] - '0')
	return (col+row)%2 == 1
}
class Solution {
public:
    bool squareIsWhite(string coordinate) {
        int col = coordinate[0] - 'a' + 1;
        int row = coordinate[1] - '0';
        return ((col + row) & 1) == 1;
    }
};
class Solution:
    def squareIsWhite(self, coordinate: str) -> bool:
        col = ord(coordinate[0]) - ord('a') + 1
        row = int(coordinate[1])
        return (col + row) % 2 == 1
var squareIsWhite = function(coordinate) {
  const col = coordinate.charCodeAt(0) - 97 + 1;
  const row = coordinate.charCodeAt(1) - 48;
  return ((col + row) & 1) === 1;
};

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